Redistribution of Charge on Conducting Plates: Calculating final charge distribution on the various faces of multiple parallel metallic plates after they are connected by wires or grounded.

CBSE + Competitive Practice Set

This practice set covers the redistribution of charge on the different faces of conducting plates when plates are isolated, connected by wires or grounded.

Important Rules:
1. Electric field inside a conductor in electrostatic equilibrium is zero.
2. Charge resides on the outer surfaces of a conductor.
3. Connected conductors are at the same potential.
4. Grounded conductor has zero potential.
5. For an isolated conductor, total charge is conserved.

E_{inside conductor} = 0

V_{connected conductors} = Same

V_ground = 0

Section A — 1 Mark Questions
10 MCQs

Q1. A conductor in electrostatic equilibrium has electric field inside it equal to: 1 Mark

A) Maximum

B) Zero

C) Infinite

D) Variable

Correct Answer: B) Zero

Solution:
Free charges inside a conductor move until the internal electric field becomes zero.

Therefore,

E = 0

Q2. If an isolated neutral conducting plate is placed near a positively charged plate, the near surface of the neutral plate acquires: 1 Mark

A) Positive charge

B) Negative charge

C) Zero charge

D) Infinite charge

Correct Answer: B) Negative charge

Solution:
Positive charge attracts electrons toward the nearer surface of the neutral conductor. Hence the near surface becomes negatively charged.

Q3. A conductor connected to the Earth is said to be: 1 Mark

A) Insulated

B) Grounded

C) Polarized

D) Neutralized completely

Correct Answer: B) Grounded

Solution:
A conductor connected to the Earth is called a grounded or earthed conductor. Its potential is taken as zero.

Q4. Two conducting plates connected by an ideal wire are necessarily at: 1 Mark

A) Different charge

B) Same potential

C) Same charge

D) Zero charge

Correct Answer: B) Same potential

Solution:
Since charge can freely flow through the connecting wire, electrostatic equilibrium requires equal potential.

V₁ = V₂

Q5. The charge density on a conducting surface is given by: 1 Mark

A) σ = qA

B) σ = q/A

C) σ = A/q

D) σ = q + A

Correct Answer: B) σ = q/A

Solution:

σ = q/A

where q is charge and A is surface area.

Q6. The electric field just outside a charged conductor is: 1 Mark

A) σ/ε₀

B) σε₀

C) ε₀/σ

D) Zero always

Correct Answer: A) σ/ε₀

Solution: For a conductor,

E = σ/ε₀

just outside its surface.

Q7. If a conductor is grounded, its potential becomes: 1 Mark

A) Q

B) Infinite

C) Zero

D) Maximum

Correct Answer: C) Zero

Solution: Earth is taken as the reference potential.

V = 0

Q8. In electrostatic equilibrium, excess charge on an isolated conductor resides: 1 Mark

A) Only inside

B) On its surface

C) At the centre only

D) Uniformly in the volume

Correct Answer: B) On its surface

Solution:
Because the electric field inside a conductor is zero, excess electrostatic charge resides on its surface.

Q9. If two isolated conductors are connected by a wire, charge flows until: 1 Mark

A) Their charges become equal

B) Their potentials become equal

C) Their sizes become equal

D) Their fields become infinite

Correct Answer: B) Their potentials become equal

Solution:
Charge flows from higher potential to lower potential until electrostatic equilibrium is reached.

Q10. The SI unit of surface charge density is: 1 Mark

A) C

B) C/m

C) C/m²

D) N/C

Correct Answer: C) C/m²

Solution:

σ = q/A

Therefore its SI unit is C/m².

Section B — 2 Mark Questions
10 Competency / Numerical Questions

Q11. A neutral conducting plate is placed near a plate carrying +10 μC. What charges are induced on the two surfaces of the neutral plate? 2 Marks

Answer:
Near surface = −10 μC
Far surface = +10 μC

Solution:
The near surface must acquire −10 μC to make the electric field inside the conductor zero.

Since the conductor was initially neutral:

−10 + q = 0

Therefore:

q = +10 μC

Q12. A conductor carries a total charge of +20 μC. If −8 μC is induced on one surface, find the charge on the remaining surface. 2 Marks

Answer: +28 μC

Total charge:

q₁ + q₂ = +20 μC

Given:

q₁ = −8 μC

Therefore:

−8 + q₂ = 20

q₂ = 28 μC

Q13. Two conducting plates have potentials 20 V and 50 V. They are connected by a wire. In which direction does charge initially flow? 2 Marks

Answer: From the 50 V plate toward the 20 V plate.

Charge flows from higher electric potential to lower electric potential until both plates reach the same potential.

Q14. A conducting surface has charge 6 μC distributed uniformly over an area of 3 m². Find its surface charge density. 2 Marks

Answer: 2 μC/m²

σ = q/A

σ = 6/3 = 2 μC/m²

Q15. Why does the electric field inside a conductor become zero in electrostatic equilibrium? 2 Marks

Answer: Because free electrons move under the influence of any internal electric field. They redistribute until the internal field is completely cancelled.

Q16. A neutral conductor is placed near a negative charge. What is the nature of charge on its nearer and farther surfaces? 2 Marks

Answer:
Near surface → Positive
Far surface → Negative

The negative external charge repels electrons away from the nearer surface. Thus positive charge appears near the external negative charge.

Q17. Two identical conducting spheres carrying charges +6 μC and −2 μC are connected by a wire. Find the final charge on each sphere. 2 Marks

Answer: +2 μC on each sphere.

Total charge:

Q = 6 − 2 = 4 μC

For identical spheres, charge divides equally:

Q/2 = 4/2 = 2 μC

Q18. What happens to the charge of a conductor when it is connected to Earth? 2 Marks

Answer: Electrons can flow between the conductor and Earth until the conductor reaches zero potential.

Q19. A conductor has charge 12 μC and surface area 4 m². Find its average surface charge density. 2 Marks

Answer: 3 μC/m²

σ = q/A = 12/4 = 3 μC/m²

Q20. Why can charge redistribution occur when two conducting plates are connected by a wire? 2 Marks

Answer: Because conductors contain free electrons. These electrons can move through the connecting wire until both conductors reach the same potential.

Section C — 3 Mark Questions
10 Competitive / CBSE Numerical Questions

Q21. A conducting plate A carries +12 μC. A neutral conducting plate B is placed near it. Find the induced charges on B assuming its total charge remains zero. 3 Marks

Answer:
Near surface of B = −12 μC
Far surface of B = +12 μC

Step 1: Electric field inside B must be zero.
Therefore the near surface gets −12 μC.

Step 2: B is neutral.

q_near + q_far = 0

Step 3:

−12 + q_far = 0

Therefore:

q_far = +12 μC

Q22. Two identical conducting spheres A and B carry +8 μC and +4 μC respectively. They are connected by a wire and then separated. Find the final charge on each. 3 Marks

Answer: +6 μC on each sphere.

Total charge:

Q = 8 + 4 = 12 μC

Identical spheres share charge equally:

Q_final = 12/2 = 6 μC

Thus each sphere has +6 μC.

Q23. A conducting plate has total charge +30 μC. The charge on one surface is −10 μC. Find the charge on the other surface. 3 Marks

Answer: +40 μC

Using charge conservation:

q₁ + q₂ = 30 μC

Given:

q₁ = −10 μC

Therefore:

−10 + q₂ = 30

Hence:

q₂ = 40 μC

Q24. A surface carries charge density 5 × 10⁻⁶ C/m². Find the electric field just outside the conductor. Take ε₀ = 8.85 × 10⁻¹² C²/Nm². 3 Marks

Answer: Approximately 5.65 × 10⁵ N/C

Use:

E = σ/ε₀

Therefore:

E = (5 × 10⁻⁶)/(8.85 × 10⁻¹²)

E ≈ 5.65 × 10⁵ N/C

Q25. Three isolated conducting plates have total charges +10 μC, 0 and −10 μC. Explain what condition must be satisfied inside each conducting plate at equilibrium. 3 Marks

Answer: The electric field inside the material of every conducting plate must be zero.

At electrostatic equilibrium, free charges stop having any net motion. Therefore:

E_{inside each conductor} = 0

The surface charges redistribute themselves to satisfy this condition while the total charge of each isolated conductor remains conserved.

Q26. Two identical spheres have charges +10 μC and −4 μC. They are connected by a wire. Find the final charge on each sphere. 3 Marks

Answer: +3 μC on each sphere.

Total charge:

Q = 10 − 4 = 6 μC

Since spheres are identical:

Q_each = 6/2 = 3 μC

Q27. A neutral conductor is placed near a positively charged conductor. Does the neutral conductor acquire a net charge? Explain. 3 Marks

Answer: No. If it remains isolated, its net charge remains zero, but its charge distribution changes.

The positive external charge attracts electrons toward the nearer surface. Hence the near surface becomes negative and the farther surface becomes positive.

The two induced charges are equal and opposite, so:

Q_net = 0

Q28. A conductor connected to Earth is placed near a positive charged body. What is the effect of grounding? 3 Marks

Answer: Electrons can flow from Earth to the conductor, making the conductor negatively charged depending on the configuration.

A positive external charge attracts electrons. Since the conductor is connected to Earth, electrons can be supplied from the Earth. The conductor remains at:

V = 0

Q29. Explain why the charges on different faces of a conducting system are not necessarily equal. 3 Marks

Answer: Surface charge distribution depends on the geometry, neighbouring charges, connections and grounding conditions.

Charges redistribute so that:

E_{inside conductor} = 0

Therefore, more charge can accumulate on a surface closer to another charged body or on a surface with suitable geometry.

Q30. Three conducting plates are connected together by conducting wires. What two important conditions are used to determine their final charge distribution? 3 Marks

Answer: 1. All connected plates must have the same potential. 2. The electric field inside each conductor must be zero.

For connected conductors:

V₁ = V₂ = V₃

For electrostatic equilibrium:

E_inside = 0

If the complete system is isolated, total charge is also conserved:

Q_initial = Q_final

These equations are used to determine the final charge on the individual faces.

Redistribution of Charge on Conducting Plates
CBSE + JEE/Competitive Practice | 30 Questions with Answers & Solutions