Gauss's Law
EXERCISES, Q-21 to Q-24
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 conducting plate = σ/εₒ.
Force on the electron in this field =σe/εₒ
For the electron motion between the plates, 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 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 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 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, 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 it. 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 plates will be due to plate X only.
Here the elctric field = σ/εₒ
= (Q/2A)/εₒ
= Q/2Aεₒ.
Towards the right of 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 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 also there are 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
Links to the Chapters
CHAPTER- 30- Gauss's Law
CHAPTER- 29- Electric Field and Potential
CHAPTER- 28- Heat Transfer
OBJECTIVE -I
CHAPTER- 26-Laws of Thermodynamics
CHAPTER- 25-CALORIMETRY
Questions for Short Answer
OBJECTIVE-I
OBJECTIVE-II
EXERCISES - Q-11 to Q-18
CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 30- Gauss's Law
CHAPTER- 29- Electric Field and Potential
CHAPTER- 28- Heat Transfer
CHAPTER- 26-Laws of Thermodynamics
CHAPTER- 25-CALORIMETRY
Questions for Short Answer
OBJECTIVE-I
OBJECTIVE-II
CHAPTER- 24-Kinetic Theory of Gases
CHAPTER- 23 - Heat and Temperature
CHAPTER- 21 - Speed of Light
CHAPTER- 20 - Dispersion and Spectra
CHAPTER- 19 - Optical Instruments
CHAPTER- 18 - Geometrical Optics
CHAPTER- 17 - Light Waves
CHAPTER- 16 - Sound Waves
CHAPTER- 15 - Wave Motion and Waves on a String
CHAPTER- 14 - Fluid Mechanics
CHAPTER- 13 - Fluid Mechanics
CHAPTER- 12 - Simple Harmonic Motion
CHAPTER- 11 - Gravitation
CHAPTER- 10 - Rotational Mechanics
CHAPTER- 9 - Center of Mass, Linear Momentum, Collision
CHAPTER- 8 - Work and Energy
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CHAPTER- 7 - Circular Motion
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CHAPTER- 6 - Friction
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CHAPTER- 6 - Friction
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CHAPTER- 5 - Newton's Laws of Motion
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CHAPTER- 4 - The Forces
The Forces-
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CHAPTER- 3 - Kinematics - Rest and Motion
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CHAPTER- 2 - "Physics and Mathematics"
CHAPTER- 2 - "Physics and Mathematics"
