AP Physics C – Electrostatics – Multiple Choice Practice Questions on Capacitors

Let us discuss some practice questions on capacitors meant for AP Physics C aspirants. You should try to find the answers yourself and then only go through the solution part. Here are the questions:

(1) A parallel plate air capacitor with separation d between the plates negligibly small compared to the length and breadth of the plates, is fully charged by connecting it across a battery of emf V volt. If the capacitance of the capacitor is C and the area of each plate is A, the electric field at a point P in between the plates, at distance d/4 from the positive plate (Fig.) is

(a) CV/ε₀A

(b) 4CV/ε₀Ad

(c) CV/ε₀Ad

(d) CV/ 4ε₀Ad

(e) 3CV/4ε₀A

The point P being sufficiently away from the edge of the plate, the electric field at P is constant and is independent of the distance from the plate. The electric field due to a plane sheet of charge is σ/2ε₀ where σ is the surface charge density (charge per unit area). [This can be easily obtained from Gauss’s law].

In the region between the plates, the fields due to the positive plate and the negative plate act in the same direction ( from positive plate to negative plate) and hence they add up to produce a net field of magnitude 2×σ/2ε₀ = σ/ε₀.

Since σ = Q/A and Q = CV, the field at P is CV/ε₀A.

[Note also that at a point outside, the electric field is zero since the positive and negative charges will produce equal and opposite fields].

(2) Two capacitors C₁ and C₂ are identical in all respects except for the dielectric media between their plates. C₁ has air as dielectric where as C₂ has a medium of dielectric constant K in place of air. The capacitor C₁ is charged to V₁ volt and the charging battery is disconnected. Then the uncharged capacitor C₂ is connected across C₁. If the common voltage across the capacitors is V₂, the value of the dielectric constant K is

(a) V₁/ V₂

(b) V₂/ V₁

(c) (V₁ + V₂)/ (V₁ – V₂)

(d) (V₁ – V₂)/ V₁

(e) (V₁ – V₂)/ V₂

If the capacitance of C₁ is C the capacitance of C₂ is KC.

Since the charge is conserved, the initial charge (Q) on C₁ must be equal to the sum of the final charges on C₁ and C₂. Therefore we have

Q = CV₁ = CV₂ + KCV₂

This gives K =(V₁ – V₂)/ V₂

(3) Two thin metal plates of the same area are given positive charges Q₁ and Q₂ and kept parallel to each other to form a capacitor of capacitance C. If Q₂ > Q₁ what will be the potential difference between the plates?

(a) (Q₂ – Q₁) /2C

(b) (Q₂ + Q₁) /2C

(c) (Q₂ + Q₁) /C

(d) (Q₂ – Q₁) /C

(e) (Q₂ + Q₁) /4C

This question may mislead many of you to a wrong answer…

The potential difference (V) between the plates is related to the electric field (E) between the plates and the distance (d) between the plates as

E = V/d

The plates produce opposing fields of magnitudes E₁ (due to Q₁) and E₂ (due to Q₂) in the region between the plates and the resultant field is directed from the plate carrying greater positive charge Q₂ to that carrying smaller positive charge Q₁.

Therefore, E = E₂ – E₁ = V/d so that

V = (E₂ – E₁)d

But E₁ = σ₁ /2ε₀ and E₂ = σ₂ /2ε₀ where σ₁ and σ₂ are the surface densities of charges on the plates given by σ₁ = Q₁/A and σ₂ = Q₂/A

Substituting these values, V = [(Q₂/2ε₀A) – (Q₁/2ε₀A)] /d

Since ε₀A/d = C, we obtain V = (Q₂ – Q₁) /2C.

(4) A parallel plate capacitor with air as dielectric has plates of area A and separation d. It is charged to a potential difference V. The charging battery is then disconnected and the plates are pulled apart so that the separation becomes 3d. What is the work done for pulling the plates?

(a) ε₀AV²/2d

(b) ε₀AV²/3d

(c) ε₀AV²/d

(d) 3ε₀AV²/d

(e) 2ε₀AV²/3d

It is enough to find the difference between the energies of the capacitor. The potential difference between the plates will increase on pulling the plates apart; but the charge will remain unchanged. The energy (U₁) before pulling is given by

U₁ = Q²/2C where C is the initial value of capacitance given by C = ε₀A/d.

When the separation between the plates is increased to 3d, the capacitance is reduced to C/3 and the energy (U₂) is given by

U₂ = Q²/(2C/3) = 3Q²/2C

The work done (W) for pulling the plates apart is given by

W = U₂ – U₁ = 3Q²/2C – Q²/2C = Q²/C

Since Q = CV where C = ε₀A/d, we have

W = CV² = ε₀AV²/d

You will find some useful multiple choice questions with solution at physicsplus.