Friday, May 29, 2026

H C Verma solutions, Chapter-30, Gauss's Law, Concepts of Physics, Part-II

Gauss's Law


Questions for Short Answer


   1. A small plane area is rotated in an electric field. In which orientation of the area is the flux of the electric field through the area maximum?  



Answer: The flux (Δⲫ) of an electric field (E) through an area (Δs) is given as,

Δⲫ = E.Δs

     =E*Δs*cosß

Where ß is the angle between the direction of the electric field and the direction of the normal to the area Δs. At a fixed place, E is constant, and the plane area Δs is also fixed. So the flux will be maximum for the maximum cosß. The maximum value of cosß is 1 for ß = 0°. That is when the direction of E and the normal to the area are the same. In other words, the flux is maximum when the plane of the area is perpendicular to the direction of the electric field.       


 

   2. A circular ring of radius r made of a nonconducting material is placed with its axis parallel to a uniform electric field. The ring is rotated about a diameter through 180°. Does the flux of the electric field change? If yes, does it decrease or increase?  



Answer: Δs = πr², 

In the first case, the axis of the ring is parallel to the direction of the electric field. Since one side of the normal to the area is assumed positive, let us assume that the normal along the electric field is positive. So in the first case, the flux of the electric field through the ring is,

 ΔΦ =E*Δs*cos0° =E(πr²) =πr²E.

When the ring is rotated through 180°, the angle between the electric field and the normal becomes 180°. Now the flux is,

 ΔΦ' =E*Δs*cos180° =-πr²E.

So the flux of the electric field changes, and it decreases. 



 

   3. A charge Q is uniformly distributed on a thin spherical shell. What is the field at the center of the shell? If a point charge is brought close to the shell, will the field at the center change? Does your answer depend on whether the shell is conducting or nonconducting? 



Answer: Since the charge is uniformly distributed on the thin spherical shell, the field vectors at the center will be symmetrical all around and cancel out. So the net field at the center will be zero.

     When a point charge is brought near the shell, we have two conditions, 

     First, when the shell is conducting. A large number of free electrons in the conductor will rearrange themselves, and the shell will no longer remain uniformly charged. But the electric field inside the conductor will be zero everywhere, even when it is in the field of a point charge, and all the charges on the conductor will be on the outer surface. If we take the Gaussian surface through the thickness of the shell, the electric field at all the points of this surface = zero. So flux through this surface is zero, and there is no charge inside this surface. So the electric field inside this surface, and hence at the center, is also zero.

      The second condition is when the shell is non-conducting. In this case, there are no large numbers of free electrons to rearrange, and there is no certainty of the electric field being zero inside the material. Hence, the shell will still remain uniformly charged, and the field due to it at the center will be zero, but the field at the center due to the point charge (q) will be non-zero and equal to kq/r².         



 

   4. A spherical shell made of plastic contains a charge Q distributed uniformly over its surface. What is the electric field inside the shell? If the shell is hammered to deshape it without altering the charge. Will the field inside be changed? What happens if the shell is made of metal?  



Answer: Just like in gravitation, the electric field due to a uniformly charged thin spherical shell at an internal point is zero. 

    When the non-conducting plastic shell is deformed, the charge distribution on it is disturbed, though the total charge is the same. Since it is non-conducting, the field inside its material may not be zero, nor will the charge reside only on the outer surface. In such a case, the field inside will not be zero, and it will change.    

     When the shell is made of metal, it is conducting. There will be large numbers of free electrons, and the field inside the metal will be zero. And by taking the Gaussian surface through the metal thickness, we can prove that the field inside will not change and will remain zero.



 

   5. A point charge is placed in a cavity in a metal block. If a charge Q is brought outside the metal, will the charge q feel an electric force?  



Answer: Due to the large number of free electrons in the metal block, they rearrange themselves such that the electric field inside the metal is always zero. If we choose a Gaussian surface S such that it is always inside the metal and enclosing cavity, the total charge inside it must be zero.


 Since there is already a charge q in the cavity, a -q charge must have been induced inside the surface of the cavity. So whatever the charge outside the metal, there will be zero electric fields in the metal body, and the charge q will not be affected.     


 

   6. A rubber balloon is given a charge Q distributed uniformly over its surface. Is the field inside the balloon zero everywhere if the balloon does not have a spherical surface?  



Answer: A rubber balloon is non-conducting. When the shape is non-spherical, the symmetry is lost. Also, there is an absence of a large number of free electrons to make the inside of the material of the balloon neutral. So the field inside the balloon will not be zero everywhere.      


 

   7. It is said that any charge given to a conductor comes to its surface. Should all the protons come to the surface? Should all the electrons come to the surface? Should all the free electrons come to the surface?   



Answer: As we know, an atom has a nucleus made of protons and neutrons, and the electrons revolve around this nucleus in different orbits. In a neutral atom, the number of protons and electrons is exactly the same. In a conductor, the atoms or the molecules do not move freely, nor do the protons in the nucleus. The electrons in the outermost orbit are weakly bonded to the nucleus compared to the inner orbit electrons.

   When a negative charge is given to the conductor, a lot of free electrons are transferred to it. Only these free electrons come to the surface. 

    When a positive charge is given to the conductor, a lot of weakly bonded electrons are taken away from it. There is a deficiency of electrons in the conductor. This electron deficiency appears on the conductor's surface. 

======================

OBJECTIVE - I


   1. A charge Q is uniformly distributed over a large plastic plate. The electric field at a point P close to the center of the plate is 10 V/m. If the plastic plate is replaced by a copper plate of the same geometrical dimensions and carrying the same charge Q, the electric field at the point P will become

(a) zero

(b) 5 V/m

(c) 10 V/m

(d) 20 V/m     



Answer: (c) 


EXPLANATION: There will be no change. Since the charge is uniformly distributed in both cases, the charge enclosed within the cylindrical Gaussian surface will also be the same. So the electric field will also be the same.       





   2. A metallic particle having no charge is placed near a finite metal plate carrying a positive charge. The electric force on the particle will be

(a) towards the plate

(b) away from the plate

(c) parallel to the plate

(d) zero      



Answer: (a) 


EXPLANATION: Since the particle is metallic, there are some loose electrons in it. Due to the positive charge on the nearby plate, the loose electrons in the particle come towards the plate, and there is a deficiency of electrons on the farther side. Though the particle as a whole is neutral, it is electrically polarized. The negative charge is near the plate, and an equal amount of positive charge is away from the plate. Thus, there is an attractive force between the plate and the negative charge on the particle, and a repulsive force between the plate and the positive charge on the particle. Though the amount of charges involved in both forces is the same, the distances between the charges are not the same. The positive charge on the metal particle is farther from the plate than the negative charge. Since the electric force is inversely proportional to the square of the distance, the attractive force is stronger than the repulsive force. So the net force is attractive and towards the plate.        





   3. A thin metallic spherical shell contains a charge Q on it. A point charge q is placed at the center of the shell and another charge q₁ is placed outside it as shown in Figure (30-Q1). All three charges are positive. The force on the charge at the center is 

(a) towards left

(b) towards the right

(c) upward

(d) zero.

The figure for Q-3


 

Answer: (d) 


EXPLANATION: If we know the field at the center of the spherical metallic shell, then the force on the charge q can be known. The field due to the charge Q at the center will be zero. The field inside the metallic shell due to the charge q₁ will be zero because the loose electrons in the metal rearrange themselves and make the field inside zero due to the outer electric field. So the net electric field at the center will be zero. Thus, the force on the charge q will be zero.       





   4. Consider the situation of the previous problem. The force on the central charge due to the shell is

(a) towards left

(b) towards the right

(c) upward

(d) zero.       



Answer: (b) 


EXPLANATION: In the steady-state, the loose electrons on the metallic shell rearrange such that they concentrate towards the charge q₁, and the opposite side gets a positive charge. Thus, the net force on the central positive charge due to the shell will be towards the right. (Mind it, the net force on the central charge due to all charges is zero)     





   5. Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10 cm surrounding the total charge is 25 V-m. The flux over a concentric sphere of radius 20 cm will be

(a) 25 V-m

(b) 50 V-m

(c) 100 V-m

(d) 200 V-m        



Answer: (a) 


EXPLANATION: Gauss's Law says that the flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by εₒ. εₒ is a constant, and the net charge enclosed by spherical surfaces of radii 10 cm and 20 cm is the same. Hence, in both cases, the flux is the same q/εₒ = 25 V-m. 





   6. Figure (30-Q2a) shows an imaginary cube of edge L/2. A uniformly charged rod of length L moves towards the left at a small but constant speed v. At t =0, the left end just touches the center of the face of the cube opposite it. Which of the graphs shown in figure (30-Q2b) represents the flux of the electric field through the cube as the rod goes through it?       


Figure for Q-6


Answer: (d) 


EXPLANATION: Suppose the total charge on the rod = q.

Charge per unit length =q/L

Length of the rod inside the cube in time t =vt.

Charge inside the cube at time t = (q/L)*vt

=(qv/L)t with a maximum limit of q/2.

This limit of q/2 will start at time L/2v when half of the rod just enters the cube. And this limit q/2 will remain the same until time L/v when the end of the rod enters the cube. After this withdrawal of the rod begins, and the charge remaining inside is = 3q/2 -(qv/L)t.  

The flux of the electric field through the cube =Q/εₒ

Between time t = 0 to t =L/2v,

Flux =(qv/Lεₒ)t

So the flux is directly proportional to the time, the graph will be a straight line starting from the origin, and the slope =qv/Lεₒ.

Between time t =L/2v to t =L/v, the flux is =q/2εₒ =constant. So the graph will be a straight line parallel to the time axis. 

Between time t =L/v to t =3L/2v, the flux will be =3q/2εₒ-(qv/Lεₒ)t. It is also a straight line graph with slope = -qv/Lεₒ. 

  The graphs a, b, and c shown in the figure are all straight lines. Only the graph d fulfills our explanations. Hence, the option (d). 





   7. A charge q is placed at the center of the open end of a cylindrical vessel (figure 30-Q3). The flux of the electric field through the surface of the vessel is

(a) zero

(b) q/εₒ

(c) q/2εₒ

(d) 2q/εₒ 


Figure for Q-7


Answer: (c) 


EXPLANATION: Assume an identical inverted cylindrical vessel over the given one.
Figure for Q-7


Now take both combined as a Gaussian surface, which is a closed cylindrical vessel. Now the charge q is inside it, just at the center. The flux through the whole closed cylindrical surface = q/εₒ.

   Hence, from symmetry, we can say that the flux of the electric field through the surface of the given open vessel = q/2εₒ.  

===================

OBJECTIVE - II


   1. Mark the correct options:

(a) Gauss law is valid only for symmetrical charge distributions.

(b) Gauss's law is valid only for charges placed in a vacuum.

(c) The electric field calculated by Gauss's law is the field due to the charges inside the Gaussian surface.

(d) The flux of the electric field through a closed surface due to all the charges is equal to the flux due to the charges enclosed by the surface.            

 

Answer: (d) 


EXPLANATION: Gauss Law is valid for symmetrical as well as non-symmetrical charge distributions. There is no condition for Gauss law to be true only in a vacuum. Hence, the options (a) and (b) are wrong. 

      Gauss law refers to the net electric field, which is the electric field due to inside charges as well as outside charges of a Gaussian surface. So the option (c) is also not true.

    Since the solid angle subtended by a closed surface at an outer point is zero, the flux of all outer charges through the closed Gaussian surface is zero. Hence, the flux of the electric field through a closed surface due to all charges is equal to the flux due to the charges enclosed by the surface. Option (d) is correct.    

 

   2. A positive point charge Q is brought near an isolated metal cube.

(a) The cube becomes negatively charged.

(b) The cube becomes positively charged.

(c) The interior becomes positively charged, and the surface becomes negatively charged.

(d) The interior remains charge-free, and the surface gets a nonuniform charge distribution.                 



Answer: (d) 


EXPLANATION: There is no change in the total charge of the metal cube due to the positive charge. Since there are numerous loose electrons in a metal piece, these loose electrons rearrange themselves to make the electric field inside the metal due to the outer charge to zero. Hence, the options (a), (b), and (c) are not true. 

     Due to the rearrangement of loose electrons, the charges come to the surface of the metal cube and polarize. The interior becomes charge-free, but the surface gets a nonuniform charge distribution. Option (d) is correct.        


 

   3. A large nonconducting sheet M is given a uniform charge density. Two uncharged small metal rods A and B are placed near the sheet as shown in Figure (30-Q4). 

(a) M attracts A

(b) M attracts B

(c) A attracts B

(d) B attracts A.         
The figure for Q-3
 

Answer: All 


EXPLANATION: The direction of the electric field due to the sheet near the blocks will be horizontal. Its direction will depend on the nature of the charge on the sheet. Suppose the charge on the sheet is positive, then the direction of the field will be from the sheet towards the blocks as shown in the figure. 
The figure for Q-3


    The loose electrons on the metal blocks will rearrange themselves to make the electric field inside equal and opposite to the field due to the sheet. For this, the ends of the blocks A and B facing the sheet will be negatively charged, while the ends away from the sheet will get positively charged (please see the figure). Since the negatively charged ends of the blocks are nearer to the positively charged sheet, M attracts both A and B. Between the blocks, the nearer ends have the opposite charges. Hence, both will attract each other. Thus, all options are true.      


 

   4. If the flux of the electric field through a closed surface is zero, 

(a) The electric field must be zero everywhere on the surface

(b) The electric field may be zero everywhere on the surface

(c) The charge inside the surface must be zero

(d) The charge in the vicinity of the surface must be zero.           



Answer: (b), (c). 


EXPLANATION: From the Gauss law, the flux through a closed surface,

Ⲫ = q/εₒ, where q is the net inside charge.

Given that Ⲫ = 0, hence q = 0.

So the net inside charge must be zero. Option (c) is true. 

   Here, the net charge inside must be zero, but there may be different positive and negative charges distributed unevenly, the sum of which is zero. So the electric field may not be zero everywhere on the surface, and there may be a charge inside in the vicinity of the surface. The options (a) and (d) are not true.

           But the electric field everywhere may be zero on the surface if there is no charge inside. Option (b) is true.   


 

   5. An electric dipole is placed at the center of a sphere. Mark the correct options.

(a) The flux of the electric field through the sphere is zero.

(b) The electric field is zero at every point of the sphere.

(c) The electric field is not zero anywhere on the sphere.

(d) The electric field is zero on a circle on the sphere.           



Answer: (a), (c). 


EXPLANATION: Since the total charge of a dipole is zero, the flux of the net electric field through the sphere is zero. Option (a) is correct. 

    The electric field at a distance r (radius of the sphere) from the center of a dipole, 

   E = (1/4πεₒ)(p/r³)√(3 cos²θ +1).  Where theta is the angle between the radius and the axis of the dipole.

It shows that the electric field is not zero at every point of the sphere. Option (b) is not true.    

       If we equate the above value of E to zero, then 3 cos²θ +1 =0

→cos²θ = -1/3, which gives cosθ value as imaginary. So the electric field is not zero at any point on the sphere. Option (c) is correct, and (d) is not true.        

 


   6. Figure (30-Q5) shows a charge q placed at the center of a hemisphere. A second charge Q is placed at one of the positions A, B, C, and D. In which position(s) of this second charge, the flux of the electric field through the hemisphere remains unchanged?

(a) A

(b) B

(c) C

(d) D.              
The figure for Q-6



Answer: (a), (c). 


EXPLANATION: Suppose the area of the surface of the hemisphere =ΔS. The solid angle subtended by this area at a point = Δ𝛀.

    Now the flux of the electric field through ΔS due to the charge at any point is given as,

  ΔⲪ =(Q/4πεₒ)Δ𝛀.
The diagram for Q-6


    To get the solid angle subtended by a curved surface at a point, we join the ends to the concerned point. As we see in the above figure, the solid angle Δ𝛀 has some non-zero values at the points B and D but is zero at the points A and C. So the flux of the electric field through the hemisphere by the second charge Q placed at A or C is zero, and the original flux due to the charge q at the center remains unchanged. Thus, only options (a) and (c) are true. 

 


   7. A closed surface S is constructed around a conducting wire connected to a battery and a switch (figure 30-Q6). As the switch is closed, the free electrons in the wire start moving along the wire. In any time interval, the number of electrons entering the closed surface S is equal to the number of electrons leaving it. On closing the switch, the flux of the electric field through the closed surface

(a) is increased

(b) is decreased

(c) remains unchanged

(d) remains zero.          
The figure for Q-7

 

Answer: (c), (d). 


EXPLANATION: A battery is analogous to a water pump connected to a pipeline. The pump creates pressure to make the water already in the pipe to move, but does not increase the amount of water in a section of the pipe at any instant. Similarly, the battery makes the pressure to move the free or lose electrons in the metal wire. At any section, the number of electrons is not increased. Hence, the wire is not negatively charged. Thus, on closing the switch, the charge enclosed by the closed surface S remains zero. So the flux of the electric field through the closed surface remains unchanged and zero. The options (c) and (d) are true only.      


    8. Figure (30-Q7) shows a closed surface that intersects a conducting sphere. If a positive charge is placed at point P, the flux of the electric field through the closed surface

(a) will remain zero

(b) will become positive

(c) will become negative

(d) will become undefined.           
The figure for Q-8

 

Answer: (b) 


EXPLANATION: The electric field at the sphere due to the positive charge at P will be more or less from right to left. Since the sphere is conducting, the loose electrons in it will rearrange themselves to make the electric field at the interior of the sphere to zero. To do this, the direction of the electric field due to these rearranged electrons inside the sphere should be equal and opposite to the electric field due to the positive charge at P i.e., from left to right. This can only happen when loose electrons on the sphere come towards the sphere's surface facing the point P, and this face becomes negatively charged. Since the sphere as a whole is uncharged, the face of the sphere away from the point P becomes positively charged. 
The diagram for Q-8


  The part of the surface of the sphere inside the closed surface is positively charged (see figure above). Hence, the flux through the closed surface will become positive. Option (b) is only correct.        

======================

EXERCISES


  1. The electric field in a region is given by E = (3Eₒ/5) i + (4Eₒ/5) j with Eₒ =2.0x10³ N/C. Find the flux of this through a rectangular surface of area 0.2 m² parallel to Y-Z plane. 



Answer: Since the area is parallel to Y-Z plane, the positive normal to this plane is along the x-axis. Hence, the area vector,

ΔS =(0.2 m²) i  

Electric field E = (3Eₒ/5) i + (4Eₒ/5) j 

The flux of this field through the given rectangular area,

ΔⲪ = E.ΔS

  ={(3Eₒ/5) i + (4Eₒ/5) j}.(0.2 m²) i

  =(3Eₒ/5)(0.20 m²)

  =(3*2.0x10³/5 N/C)*(0.20 m²)

  =240 N-m²/C.  


 


  2. A charge Q is uniformly distributed over a rod of length l. Consider a hypothetical cube of edge l with the center of the cube at one end of the rod. Find the minimum possible flux of the electric field through the entire surface of the cube. 



Answer: Since one end of the rod is at the center of the cube, the minimum flux will be when the minimum length of the rod is inside the cube. The minimum possible length inside the cube will be = l/2. Hence, the charge inside the cube = Q/2.

The flux of the electric field through the entire surface of the cube (according to Gauss law) =qᵢₙ/εₒ

 =(Q/2)/εₒ

 =Q/(2εₒ).

 

 


  3. Show that there can be no net charge in a region in which the electric field is uniform at all points. 



Answer: Consider an imaginary cube with an edge 'a' and placed such that a face of the cube is perpendicular to the uniform field. Except for this face and the opposite face, the remaining four faces are parallel to the electric field, and hence the flux of the electric field through these four faces = 0. 

       The area of each face of the cube = a². Assume that the x-axis is along the direction of the electric field. If the magnitude of the electric field = E, then the electric field, E = E i,
The diagram for Q-3


Area vector for one face ΔS= a² i 

and for the other face ΔS'= -a² i.              

Now the flux of the electric field,

 Ⲫ =E.ΔS +E.ΔS'

 =(E i).(a² i) +(E i).(-a² i)

 =0.

Since the cube is a closed surface, according to Gauss's law,

Flux, Ⲫ = qᵢₙ/εₒ

so, qᵢₙ/εₒ = 0

→qᵢₙ = 0.

So there is no net charge in a region where the electric field is uniform everywhere.  

 



  4. The electric field in a region is given by E = (Eₒx/l)i. Find the charge contained inside a cubical volume bounded by the surface x = 0, x = a, y = 0, y = a, z = 0 and z = a. Take Eₒ = 5x10³ N/C, l = 2 cm and a = 1 cm.  



Answer: The given electric field is, 

E =(Eₒx/l) i, it shows that the direction of the field is along the x-axis and its magnitude is not constant but varies with x. So at x = 0, E = 0 and at x =a, E =(Eₒa/l) i. 

Four surfaces y =0, y = a, z =0 and z =a are parallel to E, hence the flux of E through these surfaces = 0. The surfaces x =0 and x =a are perpendicular to E, but E = 0 at x =0. Hence, flux at this surface is zero. The total flux is the flux through the surface x = a.

The flux through the surface x =a is

Ⲫ = E.ΔS

  ={(Eₒa/l) i}Ⲫ.(a² i)

  =Eₒa³/l

  =5x10³*(0.01)³/0.02

  =0.25 N-m²/C

From Gauss's law here,

Ⲫ =qᵢₙ/εₒ

→qᵢₙ = Ⲫεₒ

  =(0.25 N-m²/C)*(8.85x10⁻¹² C²/N-m²)

 =2.2x10⁻¹² C.                 

 


 

  5. A charge Q is placed at the center of a cube. Find the flux of the electric field through the six surfaces of the cube. 



Answer: According to Gauss's law, the flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by εₒ.

   Hence, the flux of the electric field through the six surfaces of the cube =qᵢₙ/εₒ

 =Q/εₒ.               

 

 


  6. A charge Q is placed at a distance a/2 above the center of a horizontal, square surface of edge 'a' as shown in figure (30-E1). Find the flux of the electric field through the square surface. 
Figure for Q-6



Answer: Assume an imaginary cube of edge length 'a' placed just above the given square surface. Now the charge Q is at the center of the cube. From Gauss's law, the flux of the electric field through the closed surface (cube),

Ⲫ = qᵢₙ/εₒ = Q/εₒ

This flux is through six faces of the cube, each having an equal area. Hence, from symmetry, the flux of the electric field through one given square surface = Ⲫ/6 =Q/(6εₒ)   


 

 


  7. Find the flux of the electric field through a spherical surface of radius R due to a charge of 10⁻⁷ C at the center and another equal charge at a point 2R away from the center (figure 30-E2).
Figure for Q-7



Answer: The charge enclosed by the spherical surface, qᵢₙ = 10⁻⁷ C. 

From Gauss's law, the flux of the electric field through the given spherical surface 

  =qᵢₙ/εₒ

  =10⁻⁷/8.85x10⁻¹² N-m²/C 

  =1.1x10⁴ N-m²/C.               


  


  8. A charge Q is placed at the center of an imaginary hemispherical surface. Using symmetry arguments and Gauss's law, find the flux of the electric field due to this charge through the surface of the hemisphere (figure 30-E3).  
Figure for Q-8



Answer: Assume an identical inverted hemisphere over the given hemisphere. These two constitute a sphere that encloses the given charge Q, which is at its center. From Gauss's law, the flux of the electric field through the sphere 

     =Q/εₒ
The diagram for Q-8


    Since the charge is at the center of the sphere, from symmetry, the flux in each hemisphere will be the same, say 'F'. Hence, the flux through the sphere =2F

→2F =Q/εₒ

→F =Q/(2εₒ).                     

 

 


  9. A spherical volume contains a uniformly distributed charge of density 2.0x10⁻⁴ C/m³. Find the electric field at a point inside the volume at a distance 4.0 cm from the center. 



Answer: Consider the concentric sphere having a radius, r =4.0 cm =0.04 m.

The charge enclosed in this sphere

Q =(4πr³/3)*2x10⁻⁴ C.

The electric field, E, everywhere on the surface of this sphere will be the same and normal to the surface. Hence, the flux of the electric field through the spherical surface, 

Ⲫ =∮E.ds =∮E.ds =E∮ds

From Gauss's law, this flux will be equal to Q/εₒ. So E∮ds =Q/εₒ

→E*4πr² =Q/εₒ

→E =Q/(4πεₒr²) =(1/4πεₒ)*(Q/r²)

    = (1/4πεₒ)*(4πr³/3)*2x10⁻⁴/r² 

    = r/(3εₒ)}*2x10⁻⁴

    = 0.04*2x10⁻⁴/(3x8.85x10⁻¹²)

    = 3.0x10⁵ N/C.              


  


  10. The radius of a gold nucleus (Z = 79) is about 7.0x10⁻¹⁵ m. Assume that the positive charge is distributed uniformly throughout the nuclear volume. Find the strength of the electric field at (a) the surface of the nucleus and (b) at the middle point of a radius. Remembering that gold is a conductor, is it justified to assume that the positive charge is uniformly distributed over the entire volume of the nucleus and does not come to the outer surface?  



Answer: (a) Total charge in the gold nucleus, q =79x1.6x10⁻¹⁹ C. 

r = 7.0x10⁻¹⁵ m. 

Total flux through the surface of the nucleus, Ⲫ = ∮E.ds 

     =∮E.ds {Direction of E is perpendicular to ds} 

     =E∮ds   {E is constant at the surface}

    =E*4πr²

Assuming the surface of the nucleus as a Gaussian surface, the flux through it is

= q/εₒ

Hence E*4πr²= q/εₒ

→E =q/(4πεₒr²) 

→E =(1/4πεₒ)*(q/r²) ---------- (i)

  =9x10⁹*79x1.6x10⁻¹⁹/(7x10⁻¹⁵)²

  =23.22x10²⁰ N/C

  =2.32x10²¹ N/C


(b) Let the electric field at the middle point of a radius = E'. Consider a Gaussian surface of a concentric sphere with radius r/2. The electric field at its surface will be due to the enclosed charge and will be constant everywhere on the surface. Hence total flux =E'*4π(r/2)²

=E'*πr²

From Gauss's law, this flux

E'*πr² = q'/εₒ ------ (ii)

But q' =q{(4π(r/2)³/3)/(4πr³/3)}

 =q/8

So from (ii)

E'*πr² =q/(8εₒ)

→E' =q/8πεₒr²

     =(1/4πεₒ)*(q/r²)*(1/2)

     =E/2   {from (i)}

     =1.16x10²¹ N/C. 

Gold is a conductor because the gold atom has loose electrons in the outer orbits. These electrons move freely to form a negative or positive (lack of electrons) charge in the gold metal. Their movement makes any charge come to the surface of the metal. When we move to the nucleus, it cannot be labeled as a conductor or non-conductor because no charged particle is moving. The positive charge in a nucleus is associated with the protons, which cannot be separated from it or moved. Hence, the assumption that the positive charge is uniformly distributed over the entire volume of the nucleus is justified.    



  11. A charge Q is distributed uniformly within the material of a hollow sphere of inner and outer radii r₁ and r₂ (figure 30-E4). Find the electric field at a point P a distance x away from the center for r₁ < x <r₂. Draw a rough graph showing the electric field as a function of x for 0 < x <2r₂ (figure 30-E4).     
The figure for Q-11



Answer: Charge density,

 ⍴ =Q/{4π(r₂³-r₁³)/3}

  =3Q/{4π(r₂³-r₁³)}

Charge enclosed within a sphere of radius x 

=q ={4π(x³-r₁³)/3}*3Q/{4π(r₂³-r₁³)}

=Q(x³-r₁³)/(r₂³-r₁³)

Let the electric field at a distance x = E.

Assume a concentric sphere of radius x as a Gaussian surface. From Gauss's law,

E*4πx² =Q(x³-r₁³)/{(r₂³-r₁³)εₒ}

→E =Q(x³-r₁³)/{4πεₒx²(r₂³-r₁³)}


Graph 

From x=0 to x =r₁, E = 0. 
From x =r₁ to x =r₂, E varies from zero at r₁ to E =Q/4πεₒr₂² at x =r₂.
The graph of E does not vary linearly. As the first term varies as x³/x²=x and the second term deducts a function proportional to 1/x², hence the resulting graph will be approximately concave from the top.
For x > r₂, E =Q/4πεₒx²
So the graph of E is proportional to 1/x².
 The resulting graph is nearly like this:-  
Graph for Q-11

 

 

   12. A charge Q is placed at the center of an uncharged, hollow metallic sphere of radius a. (a) Find the surface charge density on the inner surface and on the outer surface. (b) If a charge q is put on the sphere, what would be the surface charge densities on the inner and outer surfaces? (c) Find the electric field inside the sphere at a distance x from the center in the situations (a) and (b).        



Answer: (a) Due to the charge Q placed at the center of the hollow metallic sphere, a charge -Q will be induced at the inner surface and Q at the outer surface.

Hence, the surface charge density on the inner surface = -Q/(4πa²) 

On the outer surface =Q/(4πa²)


(b) When a charge q is given to the sphere, it will reside only on the outer surface, and the inner charge will still remain = -Q. Outer charge =Q+q,

                        Hence the inner charge density =-Q/(4πa²)

          And the outer charge density = (Q+q)/(4πa²)


(c) In both situations (a) and (b), the charge enclosed in a Gaussian spherical surface of radius x < a is equal to Q. From Gauss law,

E*4πx² =Q/εₒ

→E = Q/(4πεₒx²)

Where E is the electric field at a distance x from the center inside the hollow sphere. 

  

 

 


   13. Consider the following very rough model of a beryllium atom. The nucleus has four protons and four neutrons confined to a small volume of radius 10⁻¹⁵ m. The two 1s electrons make a spherical charge cloud at an average distance of 1.3x10⁻¹¹ m from the nucleus, whereas the two 2s electrons make another spherical cloud at an average distance of 5.2x10⁻¹¹ m from the nucleus. Find the electric field at (a) a point just inside the 1s cloud and (b) a point just inside the 2s cloud.      



Answer: The charge in the nucleus is due to four protons.

 So q =4*1.6x10⁻¹⁹ C

       =6.4x10⁻¹⁹ C

The charge on the two 1s electron cloud =-2*1.6x10⁻¹⁹ C 

=-3.2x10⁻¹⁹ C 


(a) From the Gauss law, the electric field just inside the 1s electron cloud =q/4πεₒr², where q is the charge enclosed by a Gaussian spherical surface just inside the 1s cloud =The charge of the nucleus. r is the radius of this Gaussian sphere, which is r = 1.3x10⁻¹¹ m.

Hence E=q/4πεₒx²

 =(1/4πεₒ)*q/x²

 =9x10⁹*6.4x10⁻¹⁹/(1.3x10⁻¹¹)²

 =3.4x10¹³ N/C 


(b) For a point just inside the 2s electron cloud, charge enclosed =Charge of the nucleus - charge of 1s electron cloud

q'=6.4x10⁻¹⁹ -3.2x10⁻¹⁹ C

 =3.2x10⁻¹⁹ C.

Hence, the electric field just inside the 2s electron cloud,

E' =(1/4πεₒ)*q'/x²

 =9x10⁹*3.2x10⁻¹⁹/(5.2x10⁻¹¹)² 

 =1.06x10¹² N/C.        


  


   14. Find the magnitude of the electric field at a point 4 cm away from a line charge of density 2x10⁻⁶ C/m.      


Answer: The magnitude of the electric field at a distance r from the line charge,

E =λ/(2πεₒr)

where λ =linear charge density

   =2x10⁻⁶ C/m

r =4 cm =0.04 m

Hence E =λ/(2πεₒr) 

   =2λ/{(4πεₒ)r}

  =2*2x10⁻⁶*9x10⁹/0.04 N/C

  =9x10⁵ N/C. 


 

 

   15. A long cylindrical wire carries a positive charge of linear density 2.0x10⁻⁸ C/m. An electron revolves around it in a circular path under the influence of the attractive electrostatic force. Find the kinetic energy of the electron. Note that it is independent of the radius.      



Answer: Suppose the electron revolves in a circle of radius r. The electric field at distance r from the wire, E =λ/2πεₒr

Attractive force on the electron, F=qE

=λq/2πεₒr

Let the velocity of the electron = v.

So, mv²/r =λq/2πεₒr

→½mv² =λq/(4πεₒ)

→K.E. =½mv² 

   =2x10⁻⁸*1.6x10⁻¹⁹*9x10⁹ J

  =2.88x10⁻¹⁷ J. 


  

   16. A long cylindrical volume contains a uniformly distributed charge of density ⍴. Find the electric field at a point P inside the cylindrical volume at a distance x from its axis (figure 30-E5).   
The figure for Q-16

   


Answer: Consider a coaxial cylindrical Gaussian surface of radius x. The electric field, E, on the surface of this cylindrical surface will be perpendicular to the curved surface and the same everywhere.

 Charge enclosed in this surface, 

q =⍴*πx²l

Hence, from Gauss law, E*ds =q/εₒ

    Here, we will take ds = area of the curved surface only, as the two flat surfaces of the cylinder will have an electric field parallel to them. 

→E =q/2πxlεₒ 

  =ρπx²l/2πxlεₒ

  =ρx/2εₒ.  


  

   17. A nonconducting sheet of large surface area and thickness d contains a uniform charge distribution of density ⍴. Find the electric field at a point P inside the plane, at a distance x from the central plane. Draw a qualitative graph of E against x for 0<x<d.       



Answer: Consider a plate of area A and thickness 2x at the middle of the sheet as shown in the figure. Take it as a Gaussian surface.
Diagram for Q-17


    The volume of this plate =2xA. Charge enclosed in the plate q =2xA⍴. Since the sheet is large, the direction of the electric field at the surface of the plate will be perpendicular to the area A everywhere and constant = E (say). The electric field directions at the edge surfaces will be parallel. Hence, ds =2A for the whole Gaussian surface. From Gauss law, E*ds = q/εₒ

→E =2xA⍴/2Aεₒ =⍴x/εₒ.


Graph:-

The electric field E varies linearly for x = 0 to x = d/2. It is a straight line. At x =d/2, E =⍴d/2εₒ. For x > d/2, the charge enclosed in the Gaussian surface remains the same as at x=d/2. So the field remains constant equal to ⍴d/2εₒ. The graph will be as follows.    
Graph for Q-17

 

 

   18. A charged particle having a charge of -2.0x10⁻⁶ C is placed close to a nonconducting plate having a surface charge density of 4.0x10⁻⁶ C/m². Find the force of attraction between the particle and the plate.      



Answer: The electric field near a charged sheet, E = σ/2εₒ.

Here σ = 4.0x10⁻⁶ C/m²

Charge on the particle, q =-2.0x10⁻⁶ C.

The magnitude of the electric force of attraction on the particle = qE

=2.0x10⁻⁶*4.0x10⁻⁶/(2*8.85x10⁻¹²) N

=0.45 N.             

  

   19. One end of a 10 cm long silk thread is fixed to a large vertical surface of a charged nonconducting plate, and the other end is fastened to a small ball having a mass of 10 g and a charge of 4.0x10⁻⁶ C. In equilibrium, the thread makes an angle of 60° with the vertical. Find the surface charge density on the plate.       



Answer: Since the ball is away from the vertical surface, the nature of the charge on the surface is the same as on the ball.

 The horizontal force on the ball, F =qE.

  =qσ/2εₒ, where sigma is the charge density on the surface.

Weight of the ball, W =mg.

Let the tension on the string =T.
Diagram for Q-19


Now T.cos30° = F, and

T.sin30° =W

Dividing, we get,

tan30° =W/F

→F =W/tan30°

→qσ/2εₒ =mg√3

→σ =2√3mgεₒ/q

 =2√3*0.01*9.8*8.85x10⁻¹²/4x10⁻⁶ C/m²

=7.5x10⁻⁷ C/m²




 

   20. Consider the situation of the previous problem. (a) Find the tension in the string in equilibrium. (b) Suppose the ball is slightly pushed aside and released. Find the time period of the small oscillations.      



Answer: (a) In equilibrium,

T.sin30° = mg

→T = 2mg

   = 2*0.01*9.8 N

   = 0.196 N ≈ 0.20 N.


(b) The ball and thread will oscillate like a pendulum. But the value of g will be replaced with the acceleration of the ball along the thread in equilibrium under the force of tension (as if it were free to move), i.e., g' = T/m

→g' =0.196/0.01 = 19.6 m/s²

{T =0.196 N, from above (a)} 

Time period =2π√(l/g')

  =2π√(0.10/19.6) s

  =0.45 s.      


  21. Two large conducting plates are placed parallel to each other with a separation of 2.00 cm between them. An electron starting from rest near one of the plates reaches the other plate in 2.00 microseconds. Find the surface charge density on the inner surfaces.     



Answer: --

        Let the surface charge density on inner surfaces =σ

The electric field by the conducting plate = σ/εₒ.

Force on the electron in this field  =σe/εₒ

For the electron motion between the plates, the initial velocity, u = 0.

Distance travelled, S =2.00 cm =0.02 m

Time taken, t =2.00 µs =2.0x10⁻⁶ 

Since S = ut +½at²

→S =½at²

→a = 2S/t²

For this acceleration, the force on the electron = ma =2mS/t²

Equating the two expressions of the force we have,

 σe/εₒ= 2mS/t²

→σ = 2mSεₒ/et²

=2*9.1x10⁻³¹*0.02*8.85x10⁻¹²/1.6x10⁻¹⁹*(2.0x10⁻⁶)²

=0.505x10⁻¹² C/m²

=5.05x10⁻¹³ C/m².   

  


  22. Two large conducting plates are placed parallel to each other, and they carry equal and opposite charges with a surface density σ as shown in Figure (30-E6). Find the electric field (a) at the left of the plates, (b) in between the plates, and (c) at the right of the plates.     
The figure for Q - 22



Answer: As shown in the figure, the charge is spread only on the inner surfaces. Hence, the fields near them will be as if due to a sheet of charge. 

   The magnitude of the electric field by each sheet = σ/2εₒ.

       The direction of the electric field due to the left sheet will be towards it at a point, while it will be away from the right sheet.

       (a) At a point left of the plates, the direction of the electric field due to two charged sheets will be equal and opposite. Hence, the magnitude of the electric field at this point, A = zero.
Diagram for Q-22


        (b) At a point in between the plates, the direction of the electric field due to each of the charged sheets will be the same, i.e., from right to left. Hence, net electric field = σ/2εₒ+σ/2εₒ

 =σ/εₒ.  Point B.


       (c) At a point right of the plates, the direction of the electric fields due to each of the charged sheets will be equal and opposite. Hence, the net electric field at this point, C =zero.  


  

   23. Two conducting plates X and Y, each having a large surface area A (on one side), are placed parallel to each other as shown in Figure (30-E7). The plate X is given a charge Q, whereas the other is neutral. Find (a) the surface charge density at the inner surface of the plate X, (b) the electric field at a point to the left of the plates, (c) the electric field at a point in between the plates, and (d) the electric field at a point to the right of the plates.     
The figure for Q - 23



Answer: (a) One side surface area of the plate X = A. 

The total area of both sides =2A.

            When a charge Q is given to the plate X, it will spread to both sides of the plate because the charge will reside on the outer surface of the conductor. So the charge Q is equally spread on both sides of the plate. Thus, the charge at the inner surface = Q/2. 

Hence, the surface charge density at the inner surface, σ =(Q/2)/A = Q/2A.


(b) The electric field near a conducting plate = σ/εₒ

     Here, the electric field at a point to the left of the plates =(Q/2A)/εₒ

                      =Q/2Aεₒ.

For a positive Q, the direction of the field will be towards the left.


(c) The second plate is neutral, i.e., it has no charge. But due to the electric field near plate X, a charge will be induced in plate  Y. But the charge induced on the inner and outer surfaces will be of different natures, so that the inside of the material is neutral, and charges will be equal and opposite. Thus, the surfaces of plate Y will be like two charged sheets. The fields due to these two sheets will be equal and opposite and have no influence at a point near them. So the electric field at a point in between the plates will be due to plate X only. Here, the electric field =σ/εₒ

 =(Q/2A)/εₒ

 =Q/2Aεₒ.

For the positive charge Q, the direction of the field will be towards the right.


(d) Similarly, the electric field at a point to the right of the plates will be due to plate X only. 

Here the electric field = σ/εₒ

          = (Q/2A)/εₒ

          = Q/2Aεₒ. 

Towards the right of the plates.    

           




  24. Three identical metal plates with large surface areas are kept parallel to each other as shown in Figure (30-E8). The leftmost plate is given a charge Q, the rightmost a charge -2Q, and the middle one remains neutral. Find the charge appearing on the outer surface of the rightmost plate.      
The figure for Q - 24


Answer: Since the middle plate is neutral, due to induction from the left and right plates, equal and opposite charges will appear on the surfaces so that the total charge on it is zero. Let the charges on its left surface =-q and on the right surface = q. 

       It means that there is a charge of q on the right surface of the leftmost plate. It implies that the charge on the other surface of this plate is Q-q, so the total charge on it remains Q.

       Similarly, the charge on the left surface of the rightmost plate =-q. Thus, the charge on the outer surface of this plate = -2Q+q. So the charge on this plate remains = -2Q. We have to find the charge on its outer surface, -2Q+q =?

         Consider a point P inside the middle neutral plate. The field here must be zero. We have three charged surfaces left to this point, Q-q, q, and -q. On the right, there are also three charged surfaces, q, -q, and -2Q+q, i.e., -(2Q-q). 

  If we take the direction of the field towards the right as positive, the net field at point P is,

(Q-q)/2Aεₒ+q/2Aεₒ-q/2Aεₒ-q/2Aεₒ+q/2Aεₒ+(2Q-q)/2Aεₒ=0

→Q-q+2Q-q =0

→3Q-2q =0

→q = 3Q/2

Hence, the charge on the outer surface of the rightmost plate = -2Q+q

=-2Q+3Q/2 

=(-4Q+3Q)/2

=-Q/2. 

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Links to the Chapters



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CHAPTER-6 - Friction

Questions for Short Answers

Objective - I

Objective - II

Exercises - Q 1 to Q 10

Exercises - Q 11 to Q 20

Exercises - Q 21 to Q 31

CHAPTER-7 - Circular Motion

Questions for Short Answers

Objective - I

Objective - II

Exercises - Q 1 to Q 10

Exercises - Q 11 to Q 20

Exercises - Q 21 to Q 30

CHAPTER-8 - Work and Energy

Questions for Short Answers

Objective - I

Objective - II

Exercises - Q 1 to Q 10

Exercises - Q 11 to Q 20

Exercises - Q 21 to Q 30

Exercises - Q 31 to Q 42

Exercises - Q 43 to Q 54

Exercises - Q 55 to Q 64


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CHAPTER- 9 - Center of Mass, Linear Momentum, Collision

Questions for Short Answers

Objective - I

Objective - II

EXERCISES Q-1 TO Q-10

EXERCISES Q-11 TO Q-20

EXERCISES Q-21 TO Q-30

EXERCISES Q-31 TO Q-42

EXERCISES Q-43 TO Q-54

EXERCISES Q-55 TO Q-64

CHAPTER- 10 - Rotational Mechanics

Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES Q-01 TO Q-15

EXERCISES Q-16 TO Q-30

EXERCISES Q-31 TO Q-45

EXERCISES Q-46 TO Q-60

EXERCISES Q-61 TO Q-75

EXERCISES Q-76 TO Q-86


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CHAPTER- 11 - Gravitation


Questions for Short Answers

OBJECTIVE - I

OBJECTIVE - II

EXERCISES Q-01 TO Q-10

EXERCISES Q-11 TO Q-20

EXERCISES Q-21 TO Q-30

EXERCISES Q-31 TO Q-39 (With Extra 40th problem)

CHAPTER- 12 - Simple Harmonic Motion 


Questions for Short Answers 


OBJECTIVE - I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-40

EXERCISES - Q-41 TO Q-50

EXERCISES - Q-51 TO Q-58 with EXTRA QUESTIONS Q-59 and Q-60 


CHAPTER- 13 - Fluid Mechanics 


Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-35



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CHAPTER- 14 - Some Mechanical Properties of Matter 


Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q -11 TO Q -20

EXERCISES - Q -21 TO Q -32

CHAPTER- 15 - Wave Motion and Waves on a String

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-40

EXERCISES - Q-41 TO Q-50

EXERCISES - Q-51 TO Q-57

CHAPTER- 16 - Sound Waves

Questions for Short Answers

OBJECTIVE-I

OBJECTIVE-II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-30

EXERCISES - Q-31 TO Q-40

EXERCISES - Q-41 TO Q-50

EXERCISES - Q-51 TO Q-60

EXERCISES - Q-61 TO Q-70 

EXERCISES - Q-71 TO Q-80

EXERCISES - Q-81 TO Q-89

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CHAPTER- 17 - Light Waves


CHAPTER- 18 - Geometrical Optics

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Part-II

Solutions - "Concepts of Physics" Part-II, by H C Verma

CHAPTER- 23 - Heat and Temperature

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 TO Q-10

EXERCISES - Q-11 TO Q-20

EXERCISES - Q-21 TO Q-34


CHAPTER- 24 - Kinetic Theory of Gases

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q1 to Q10

EXERCISES - Q-11 to Q-20

EXERCISES - Q-21 to Q-30


EXERCISES - Q-31 to Q-40

EXERCISES - Q-41 to Q-50

EXERCISES - Q-51 to Q-62




CHAPTER- 25 - Calorimetry

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 to Q-10

EXERCISES - Q11 to Q-18 




CHAPTER- 26 - Laws of Thermodynamics

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 to Q-10

EXERCISES - Q-11 to Q-22




CHAPTER- 27 - Specific Heat Capacities of Gases

Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 to Q-10 




CHAPTER- 28 - Heat Transfer


Questions for Short Answer

OBJECTIVE - I

OBJECTIVE - II

EXERCISES - Q-1 to Q-10








CHAPTER- 29 - Electric Field and Potential



  

















    



CHAPTER- 31 - Capacitors
























































































 





































































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