Wednesday, February 17, 2021

H C Verma solutions, Capacitors, EXERCISES, Q21-Q30 Chapter-31, Concepts of Physics, Part-II

Capacitors


EXERCISES, Q21 - Q30


   21. The particle shown in figure (31-E11) has a mass of 10 mg and a charge of -0.01 µC. Each plate has a surface area of 100 cm² on one side. What potential difference V should be applied to the combination to hold the particle P in equilibrium?  
The figure for Q-21



Answer: Mass of the particle,

 m = 10 mg =1.0x10⁻⁵ kg. 

The charge on the particle

 Q =-0.01 µC =-0.01x10⁻⁶ =-1.0x10⁻⁸ C.

The weight of the particle acts downward. The direction of the electric field is also downward between the capacitor but the charge on the particle is negative hence the electric force on the particle is upward. When these two forces are equal and opposite the particle, P will be in equilibrium. If the electric field between the capacitor plates = E, then the electric force on the particle P = QE.

For equilibrium, QE = mg

→QV'/d =mg,    

{V' is P.D. across a capacitor}

→QV'C/εₒA = mg 

{Since C =εₒA/d, →d =εₒA/C}

→V' =mgεₒA/QC  ----------- (i)

Since both capacitors have the same capacitance, the potential difference across each capacitor will be the same. Hence, V =2V', →V' =V/2. We have the values, C = 0.04 µF =4x10⁻⁸ F, A =100 cm² =0.01 m². Now (i) becomes,

V/2 =1x10⁻⁵*9.8*8.85x10⁻¹²*0.01/(1x10⁻⁸*4x10⁻⁸)

→V =2*21.6x10⁻³ V

      =43 mV. 

        




 

  22. Both the capacitors shown in figure (31-E12) are made of square plates of edge a. The separations between the plates of the capacitors are d₁ and d₂ as shown in the figure. A potential difference V is applied between points a and b. An electron is projected between the plates of the upper capacitor along the central line. With what minimum speed should the electrons be projected so that it does not collide with any plate? Consider only the electric forces. 
The figure for Q-22



Answer: The area of plates, A =a². The capacitance of the upper plate,

C' = εₒA/d₁

The capacitance of lower plate,

C" =εₒA/d₂

Equivalent capacitance, C, of these two capacitors connected in series

1/C = 1/C' +1/C"

→1/C =(d₁+d₂)/εₒA

→C =εₒA/(d₁+d₂)

Charge on the equivalent capacitor,

Q =CV

It will be the charge on each capacitor also. Hence the potential difference between the upper plate,

V' =Q/C'

  =CV/(εₒA/d₁)        

  ={εₒA/(d₁+d₂)}V/(εₒA/d₁)

  =d₁V/(d₁+d₂)

Thus the field inside the upper capacitor,
E =V'/d₁
   =V/(d₁+d₂)
Electric force on the electron F =eE
→F =eV/(d₁+d₂)
 Let the projected velocity of the electron = u
Since there is no force on it along the horizontal direction, it will travel horizontally with this uniform velocity, u.
Time taken by it to travel the plate length a is t =a/u
In this time t, the electron should cover a vertical distance = d₁/2. Since there is the electric force in the vertical direction it will be an accelerated motion. 
  Acceleration f =F/m
→f =eV/{m(d₁+d₂)}
From the accelerated motion formula, s =ut+½ft²
→d₁/2 =0 +½ft²
→t² =d₁/f
→(a/u)² =d₁*m(d₁+d₂)/eV
→u² =Vea²/{md₁(d₁+d₂)}
→u =[Vea²/{md₁(d₁+d₂)}]½ 





 


  23. The plates of a capacitor are 2.00 cm apart. An electron-proton pair is released somewhere in the gap between the plates and it is found that the proton reaches the negative plate at the same time as the electron reaches the positive plate. At what distance from the negative plate was the pair released? 



Answer: Suppose the pair is at a distance x from the negative plate. So the proton travels a distance x but the electron travels 2-x cm. The electric force on each of them is, F =eE but opposite in direction. If mass of electron is m and that of proton =m' then acceleration of the electron f=eE/m, and the acceleration of proton f' =eE/m'. Initial veleocity of each particle is zero. If they both take time t to reach the plates, then from the motion formula,

s =ut+½ft²,

For proton,

x =0 +½(eE/m')t²

→t² =2xm'/eE

For the electron,

2-x =0+½(eE/m)t²

→t² =2m(2-x)/eE

Equating these two values of t² we get,

2xm'/eE =2m(2-x)/eE

→xm' =m(2-x)

→(2-x)/x =m'/m

Now, m'/m is the ratio of the weights of a proton and electron. Since the mass of a proton is 1836 time more than the mass of an electron,

(2-x)/x =1836

→1836x =2-x

→1837x =2

→x =2/1837 cm =1.08x10⁻³ cm              






 

  24. Convince yourself that parts (a), (b), and (c) of the figure (31-E13) are identical. Find the capacitances between points A and B of the assembly.   
The figure for Q-24



Answer: All parts are identical as in each figure, 1 µF and 3 µF capacitors are in series. Similarly, 2 µF and 6 µF are in series. These two series combinations are in a parallel arrangement between points A and B. Also, the 5 µF capacitor is joined as a bridge between the two series-connected capacitors. Hence all three are identical.

    To find the capacitance of the assembly between A and B let us first remove the 5 µF capacitor and find the potential difference across it in the circuit. Equivalent capacitance of upper series, C' =3/(1+3) =3/4 µF. If a potential difference of V is applied between A and B, charge in this equivalent capacitor,

Q' =C'V =3V/4 µC, it will also be the charge on each of 1 µF and 3 µF capacitors. So potential difference across the 1 µF capacitor V' =Q'/(1 µF)

→V' = 3/4 V

Now the lower series, equivalent capacitance, 

C" =2*6/(2+6) =12/8 =3/2 µF.

Charge Q" =C"V =3V/2 µC

The potential difference across 2 µF capacitor, V" =Q"/(2 µF)

→V" =3V/(2*2) =3/4 V

Since V' = V", The potential differences across A and C, and across A and D are the same. Thus there is no potential difference across 5 µF capacitor (between C and D). See diagram (a) below.  

Diagram for Q-24


With no potential difference across 5 µF capacitor, it is ineffective. Now the assembly can be seen as two capacitors C' and C" in parallel. Diagram (b). Now the equivalent capacitance of the whole given assembly,

C =C' +C" =3/4 µF +3/2 µF

→C =(3+6)/4 µF = 9/4 µF

→C =2.25 µF.       


NOTE: In general, if such configuration of bridge assembly is there in which proportion of capacitance of upper series and lower series is same then there is no potential difference across the bridging as in the 5 µF capacitors. Here for example,

1 µF: 3 µF = 2 µF: 6 µF




 


 

  25. Find the potential difference Vₐ-Vᵦ between the points a and b shown in each part of the figure (31-E14). 
The figure for Q-25



Answer: (a) The arrangement of figure (a) can be simplified without any difference as below.        

Simplified Diagram


Equivalent capacitance between points A and B =4 +(4*2)/(4+2) =4+(4/3) =16/3 µF.

Equivalent capacitance between points A and D (series connected 16/3 µF and 2 µF) =(16/3)*2/(16/3 +2)

=(32/3)/(22/3) µF

=32/22 =16/11 µF.

Now the charge supplied by the battery,

Q =CV =(16/11)*12 µC = 192/11 µF

This charge will be on capacitor C and the equivalent capacitor of C', C" and C*. The potential difference across A and B,

V' =Q/(16/3) =3Q/16

→V' =3*192/(11*16) =36/11 volts

Equivalent capacitance of C" and C*,

Cⁿ =C"C*/(C"+C*) =8/6 µF =4/3 µF

Charge on Cⁿ, q =CⁿV' =(4/3)*(36/11) µC = 48/11 µC. 

      This same charge will be on C" and C*. Hence the potential difference across C* between a and B,

=q/C* = (48/11)/4 =12/11 V.     


(b)  Since the batteries are reverse connected, net potential difference in the circuit = 24 V - 12 V =12 V

  The capacitors are connected in series, their equivalent capacitance, C is given as  

1/C = 1/2+1/4

C =2*4/6=4/3 F

Hence the charge on the equivalent capacitor,

Q = CV =(4/3)*12 = 16 C

It is a series connection, hence this charge will be there on each capacitor. Now potential difference across 2 F capacitor,

V' =Q/C =16/2 =8 v 

This capacitor is between points a and b. Since the charge in the circuit will flow anticlockwise, point a is at a lower potential than point b. Hence, 

Va - Vb = -8  V


(c) Let us name the capacitors and batteries as blow. 

Diagram for (c)


Suppose the positive terminal of battery B supplies charge Q to the left plate of capacitor A. So the charge on the right plate = -Q. From symmetry, the negative terminal of battery B' will supply -Q charge to the right plate of the capacitor B. Hence left plate of B will have charge Q. Let the charge on the capacitor C is Q'. Since the right plate of A has charge -Q, the total charge of plates connected to it (plates of B and C) should be equal to Q. Thus Q' +Q = Q.

→Q' = 0.

So Vₐ -Vᵦ =Q'/C =0 (zero).  

                

(d) Suppose the potential at a = Vₐ = 0.

Potential at A = Vₐ+6 =6 volts.

Potential at B = Vₐ+12 =12 volts

Potential at C = Vₐ+24 =24 volts

Diagram for Q-25(d)


Let the charge on the capacitor plate near A =Q₁, so the charge on the opposite plate =-Q₁.

Similarly if the charge on the capacitor plate near B =Q₂, then the charge on the opposite plate =-Q₂. And if the charge on the capacitor plate near point C =Q₃, then the charge on the opposite plate = -Q₃. Since all the opposite plates are connected to each other only, sum of the charges on them should be zero. Vᵦ  

→ -Q₁ -Q₂ -Q₃ =0

→Q₁+Q₂+Q₃ = 0 -------- (i)

The potential difference between A and B is

 = Vᵦ-6

So, Vᵦ-6 =Q₁/4

→Q₁ =4Vᵦ -24

Similarly, between points B and b,

Q₂ =2(Vᵦ-12) 

     =2Vᵦ-24

And between points C and b
Q₃ =Vᵦ -24

Putting these values in (i) we get,
 

4Vᵦ-24 +2Vᵦ-24 +Vᵦ-24 = 0

→7Vᵦ =72

→Vᵦ =72/7 =10.3 volts

Hence,Vₐ -Vᵦ =0 -10.3 =-10.3 V


 




 


  26. Find the equivalent capacitances of the combinations shown in figure (31-E15) between the indicated points. 
The figure for Q-26



Answer: (a) Let us apply a potential difference of V between the points by a battery. 
Diagram for Q-26(a)


Suppose a charge Q is supplied by the positive terminal and -Q by the negative terminal. Q is distributed as Q' to capacitor 1 µF and Q" to 3 µF capacitor. Similarly, -Q is distributed as Q₁' to 3 µF capacitor and Q₁" to 1 µF capacitor on the right side. Let the potential at points C and D be V' and V" respectively. Between C and D, let the 4 µF capacitor gets Q* charge. All opposite plates will have opposite charges. We have,

Q₁' = 3V"

Q₁'' =1*V' =V'

Q* = 4(V"-V')

Q' =1*(V-V'') =V-V"

Q" =3*(V-V') =3(V-V')

Now, at point D, all plates are isolated, the sum of charges will be zero.

Q₁' +Q*-Q' =0 

Q' =Q₁' +Q*

→V-V" =3V"+4(V"-V')

→V-V" =7V"-4V'

→8V"-4V' = V  ----------- (i)

Also, at C, sum of charges are zero

Q₁'' -Q"-Q* =0

Q₁'' =Q"+Q*

→1*V' =3(V-V')+4*(V"-V')

→V' =3V-3V'+4V"-4V' =3V-7V'+4V"

→8V'-4V" =3V  -------- (ii)

Solving (i) and (ii) we get,

Multiplying (i) by 2 and adding to (ii),

12V" =2V+3V =5V

→V" =5V/12

From (ii),

8V'-4*5V/12 =3V

→8V' =3V+20V/12 =3V+5V/3

→8V' =14V/3

V'= 7V/12

Let the equivalent capacitance = C. Since Q =Q'+Q"

CV =1*(V-V") +3*(V-V')

→CV=V-V"+3V-3V' =4V-5V/12-3*7V/12

→CV =4V-5V/12-7V/4

→C = 4-5/12 -7/4

→C =(48-5-21)/12

→C =22/12 = 11/6 µF.

 

(b) Let us draw the diagram as below. 

Diagram for Q-26(b)


We have applied a potential difference of V between points P and N. N is connected to the negative terminal of a battery and we assume it to be at zero potential. P is connected to the positive terminal of the battery and its potential is at V. We assume the potentials of the points A, B and C as V₁, V₂ and V₃ and charge on 4 µF capacitors as Q₄ and Q₅. 

Now,

Q₁' =3V₁, Q₂' =2V₂, Q₃' =V₃ 

Q₁ =V-V₁, Q₂=2(V-V₂), Q₃=3(V-V₃)

Q₄ =4(V₁-V₂), Q₅=4(V₂-V₃)

Now, at point A, connected plates will have a total charge of zero. So, Q₁=Q₁'+Q₄ 

→V-V₁=3V₁+4(V₁-V₂)

→V =8V₁-4V₂ ----- (i) 

At point B, the total charge zero

Q₂ +Q₄=Q₅+Q₂'

→2(V-V₂) +4(V₁-V₂) =4(V₂-V₃)+2V₂

→2V-2V₂ +4V₁-4V₂=4V₂-4V₃+2V₂

→2V =-4V₁+12V₂-4V₃

→V =-2V₁+6V₂-2V₃ ----------(ii)

At point C also,

Q₃' =Q₃+Q₅

→V₃ =3(V-V₃)+4(V₂-V₃)

→V₃ =3V-3V₃+4V₂-4V₃

→3V = 8V₃-4V₂ ---------------(iii)

From (i) 4V₂ =8V₁-V, →V₂ =2V₁-V/4.

From (iii)8V₃=3V+4V₂ =3V+8V₁-V

                      =2V+8V₁ 

→V₃ =V₁+ V/4

  Putting the value of V₂ and V₃ in (ii) we get,

V =-2V₁+12V₁-3V/2-2V₁-V/2

→V+2V =8V₁ 

→V₁ =3V/8

 

V₂ =2(3V/8)-V/4

     =3V/4 -V/4

     =V/2 


And V₃ =3V/8+ V/4

            =(3+2)V/8

          =5V/8

     

Since -Q =-Q₁' -Q₂' -Q₃' we have

CV = 3V₁+2V₂+V₃

{Where C is the equivalent capacitance of the whole arrangement between points P and N.} 

→CV=3*3V/8+2*V/2+ 5V/8

→C =9V/8 +V +5V/8 µF

→C =(9+ 8+ 5)/8 µF

→C =22/8 = 11/4 µF


(c) Let us connect the indicated points to a battery that applies a potential difference of V between points P and N.

Diagram for Q-26(c)


 Through point P a charge Q is supplied by the battery and through point N a charge of -Q. N is at zero potential and P is at V potential. Suppose Q is distributed to the three connected capacitors as Q', Q" and Q*. Similarly -Q is distributed as -q, -q' and -Q*. Let the charge on 5 µF capacitor =q", and the potentials of point A and B are V' and V" respectively.

  We have, q =4V', q' =8V", Q* =4V 

Q' =2(V-V'), Q" =4(V-V"), q" =5(V'-V")

Capacitors connected to A have isolated plates. Hence the sum of charges on these plates will be zero.

-Q' + q +q" =0

→-2(V-V') + 4V' +5(V'-V") =0

→-2V+2V' +4V'+5V'-5V" =0 

→-2V +11V' -5V"=0

→11V'-5V" = 2V ---- (i)

Similarly point B is isolated

-Q" -q" +q' =0

→ -4(V-V")-5(V'-V")+8V" =0

→ -4V+4V" -5V'+5V"+8V" =0

→ 17V"-5V' =4V ------ (ii)

Multiplying (i) by 17/5 and adding to (ii),
11*17V'/5 -5V' =2*17V/5+4V
→(11*17-25)V'/5 = (34+20)V/5
→ 162 V' =54V 
→ V' = V/3
Putting this value in (ii),
17V"-5V/3 =4V
→17V" =17V/3 
→V" =V/3

Point N is also connected to isolated capacitor plates. Hence the supplied -Q charge is distributed to these plates.
-q -q' -Q* =-Q
→4V' +8V" +4V = CV
{Where C is the equivalent capacitance of the arrangement between P and N.}
→CV =4V/3 +8V/3 +4V
→  C =4/3 +8/3 +4
      =(4 +8 +12)/3
      =24/3
      =8 µC                   


(d) Let us apply a potential difference of V volts across the given points P and N. N is connected to the negative terminal of a battery and we assume its potential at zero. The potential of point P is V. If P receives a charge Q⁰ then N receives a charge of -Q⁰. 
Diagram for Q- 26(d)


       Q⁰ is distributed to connected capacitors 2 µF and 4 µF as Q and Q' respectively considering symmetry as shown in the figure. Similarly -Q⁰ is distributed to 4 µF and 8 µF capacitors as -q and -q" respectively. We take potential of A and D = V' and potential of B and C =V". Charges on capacitors between these points are Q" and Q* as shown in the diagram considering symmetry.
Now, q =4V', q' =8V", Q =2(V-V'), 
Q' =4(V-V"), Q" =6(V"-V'), 
Q* =6(V"-V") =0

Sum of charges on plates connected to point A =0.
→q+Q"-Q =0
→4V' +6(V"-V') -2(V-V') =0
→4V' +6V" -6V' -2V+2V' =0
→6V" =2V
→V" =V/3

Similarly at point B,
q'+Q*-Q'-Q" =0
→8V"+0 -4(V-V") -6(V"-V') =0
→8V" -4V+4V" -6V" +6V' =0
→6V" +6V' =4V
→6V/3 +6V' =4V
→6V' = 2V 
→V' =V/3.

              Considering the distribution of charge -Q⁰,
-Q⁰ =-2q -2q'
→CₑV =2*4V' +2*8V"
Where Cₑ is the equivalent capacitance between the given points.
→CₑV =8*V/3 +16*V/3
→Cₑ =24/3 =8 µF.   

           


 

 


  27. Find the capacitances of the combination shown in figure (31-E16) between A and B. 
The figure for Q-27


Answer: Let us draw and name some points in the diagram below. 

Diagram for Q-27


In the path GMH, two capacitors are in series. Equivalent capacitance of these two, C' is given as

1/C' = 1/2 +1/2 =1

→C' =1 µF.

Now this C' is in parallel with 1 µF capacitor between G and H. Equivalent capacitance of these two capacitors,

C" = C' +1 = 1 +1 =2 µF.

Now this C" and 2 µF between points G and E are in series. Equivalent capacitance of these two, 

1/C* = 1/C" +1/2 =1/2 +1/2 = 1          →C* = 1 µF

Between points, E and F, C* and 1 µF are in parallel {See diagram (b)}. Equivalent capacitance of these two 

C₁ =C* +1 =1 +1 =2 µF.

C₁ between E and F and 2 µF between E and C are in series. The equivalent capacitance of these two is given as 

1/C₂ =1/C₁ +1/2 =1/2 +1/2 =1 

→C₂ = 1 µF. 

Now 1 µF capacitor between points C and D, and C₂ are in parallel. The equivalent capacitance of these two 

C₃ =C₂ +1 = 1 +1 =2 µF

Between points A and B, C₃ and 2 µF are in series {Diagram (c)}. The equivalent capacitance between A and b is given 

1/Cₑ =1/C₃ +1/2 =1/2 +1/2 =1

→Cₑ = 1 µF.          




  

  28. Find the equivalent capacitances of the infinite ladder shown between points A and B. 
The figure for Q-28



Answer: Suppose the capacitance between points A and B = C. Since the series is infinite, the capacitance between points D and E on the right is also =C. Now the series becomes as in diagram (b).
Diagram for Q-28

        

    In the path ADEB, 2 µF and C are in series. Equivalent capacitance of these two is given as,

1/C' =1/2 +1/C = (C+2)/2C

→C' =2C/(C+2)

Now C' and 1 µF are in parallel. The equivalent capacitance of these two is the equivalent capacitance of the whole arrangement between A and B which we have assumed as C. Hence,

C =C' +1 =2C/(C+2) +1

→C =(2C +C+2)/(C+2)

→C = (3C+2)/(C+ 2)

→ C²+2C =3C +2

→C² -C -2 =0

→C² -2C +C -2 =0

→C(C-2) +(C-2) =0

→(C-2)(C+1) =0

Either C = 2 µF or -1 µF.   

Since negative capacitance is not acceptable, C =2 µF.







 

  29. A finite ladder is constructed by connecting several sections of 2 µF, 4 µF capacitor combinations as shown in figure (31-E18). It is terminated by a capacitor of capacitance C, such that the equivalent capacitance of the ladder between the points A and B becomes independent of the number of sections in between? 
The figure for Q-29


Answer: The given condition will be fulfilled if the equivalent capacitance of the end section becomes equal to C. Let us calculate the capacitance of the rightmost end section.

    C and 4 µF are in series. Equivalent capacitance of these two is given as,

1/C' =1/C +1/4 =(C+4)/4C

→C' =4C/(C+4)         

Now C' and 2 µF are in parallel, we have to make equivalent capacitance of these two as C. So

C' +2 =C

→4C/(C+4) +2 = C

→4C+2C+8 =C² +4C

→C²-2C-8 = 0

→C² -4C +2C -8 =0

→C(C-4) +2(C-4) =0

→(C-4)(C+2) =0

→Either C =4 µF or -2 µF. We will take the positive value, hence, C =4 µF. 






 

  30. A charge of +2.0x10⁻⁸ C is placed on the positive plate and a charge of -1.0x10⁻⁸ C on the negative plate of a parallel plate capacitor of capacitance 1.2x10⁻³ µF. Calculate the potential difference developed between the plates. 



Answer: When a simple parallel plate capacitor has equal and opposite charges on its plates (Q and -Q) we say the charge on the capacitor is Q. This Q is the net average chare, i.e. {Q-(-Q)}/2. When there are unequal charges, the net average charge on the capacitor,

q ={Q-(-Q')}/2, 

here, q ={2x10⁻⁸ -(-1.0x10⁻⁸)}/2

→q =3x10⁻⁸/2 =1.5x10⁻⁸ C.

Given capacitance C' =1.2x10⁻³ µF

→C' =1.2x10⁻⁹ F

Hence the potential difference developed between the plates,

V =q/C' 

   =1.5x10⁻⁸/1.2x10⁻⁹ V

   =15/1.2 V

   =12.5 V          

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Click here for → Friction - OBJECTIVE-II

Click here for → EXERCISES (1-10)

Click here for → Exercises (11-20)

Click here for → EXERCISES (21-31)

For more practice on problems on friction solve these- "New Questions on Friction".

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CHAPTER- 5 - Newton's Laws of Motion


Click here for → QUESTIONS FOR SHORT ANSWER

Click here for→Newton's Laws of Motion,Exercises(Q.No. 13 to 27)

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CHAPTER- 4 - The Forces

The Forces-

"Questions for short Answers"    


Click here for "The Forces" - OBJECTIVE-I


Click here for "The Forces" - OBJECTIVE-II


Click here for "The Forces" - Exercises


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CHAPTER- 3 - Kinematics - Rest and Motion

Click here for "Questions for short Answers"


Click here for "OBJECTIVE-I"


Click here for EXERCISES (Question number 1 to 10)


Click here for EXERCISES (Question number 11 to 20)


Click here for EXERCISES (Question number 21 to 30)


Click here for EXERCISES (Question number 31 to 40)


Click here for EXERCISES (Question number 41 to 52)


CHAPTER- 2 - "Physics and Mathematics"

Click here for "Questions for Short Answers"


Click here for "OBJECTIVE-II"

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