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.

AP Physics B & C – Electrostatics – Multiple Choice Questions on Capacitors

Here are some typical multiple choice questions (for practice) on capacitors:

(1) Capacitors C₁, C₂ and C₃ of values 15 μF, 10 μF, 3 μF are connected in series and the series combination is connected across a battery of emf 10 V. When the capacitors are fully charged, the charge on one plate of the 3 μF capacitor will be of magnitude

(a) 10 μC

(b) 15 μC

(c) 3 μC

(d) 20 μC

(e) 280 μC

The series combined value (C) of the three capacitors is given by

1/C = 1/C₁ + 1/C₂ + 1/C₃ = 1/15 + 1/10 + 1/3 so that C = 2 μF

The charge (Q) on this equivalent capacitance is given by

Q=CV = 2 μF ×10 V = 20 μC

When capacitors are connected in series, the charges on the plates of all capacitors are of the same magnitude so that the correct option is 20 μC [Option (d)].

(2) A 2 μF capacitor connected in a circuit has one plate at + 6 V and the other plate at – 6 V. The charge on the negative plate of the capacitor is

(a) + 12 μC

(b) – 12 μC

(c) + 24 μC

(d) – 24 μC

(e) zero

The magnitude of charge on either plate is given by

Q = CV

Note that V is the potential difference between the plates and is equal to 6 – (– 6) = 12 V.

Therefore, Q = 2 μF × 12 V = 24 μC

Since the charge on the negative plate must be negative, the answer is – 24 μC.

(3) A parallel plate capacitor with air as dielectric remains connected across a battery of emf 6 V. The charge on the capacitor in this condition is Q. If the separation between the plates is decreased by 10% in this condition and sufficient time is allowed to attain steady state, the charge on the capacitor will be

(a) unchanged

(b) increased by 9%, approximately

(c) decreased by 9%, approximately

(d) decreased by 11%, approximately

(e) increased by 11%, approximately

The correct option is (e) since the capacitance will increase by approximately 11% in accordance with the expression,

C = ε₀A/d when the separation d changes to 0.9d.

[Remember that Q = CV]

Now, note the following in connection with the above question:

If the battery is disconnected after charging the capacitor to have an initial charge Q,

(i) the charge on the plates will be unchanged if the plate separation is decreased or increased (since there is no battery to control the charge). The law of conservation of charge is very strictly obeyed.

(ii) The potential difference between the plates of the capacitor will be decreased on decreasing the separation between the plates since V = Q/C and Q is unchanged where as C is increased.

(iii) The potential difference between the plates of the capacitor will be increased on increasing the separation between the plates since V = Q/C and Q is unchanged where as C is decreased.

(iv) If a dielectric slab is inserted in to the gap between the plates, the potential difference between the plates will be decreased since the capacitance is increased (with the charge on the plates unchanged).

* * * * * * * * * * * * * *

(4) A parallel plate capacitor with air as dielectric remains connected across a battery. After the capacitor is fully charged in this condition, a slab of dielectric constant 5 is slowly introduced into the gap between the plates. Which one of the following statements is true during the introduction of the slab?

(a) The charge on the capacitor will gradually decrease

(b) The capacitance will gradually decrease

(c) A current will flow through the leads connecting the capacitor to the battery

(d) The potential difference between the plates of the capacitor will gradually increase

(e) The potential difference between the plates of the capacitor will gradually decrease

When the slab is introduced, the capacitance gradually increases and the battery supplies more charges to the capacitor since the charge Q has to increase in accordance with the equation, Q = CV. The potential difference across the capacitor will be unchanged since it is connected across the battery (whose emf is fixed).

The only correct option therefore is (c).

(5) Half of the space between the plates of a parallel plate air capacitor of capacitance C is filled as shown with a material of dielectric constant K. The new capacitance will be

(a) KC

(b) KC/2

(c) (K + 1)C/2

(d) 2C/K

(e) (K – 1)C

Initially the capacitance C is given by

C = ε₀A/d where A and d are the plate area and plate separation respectively.

On introducing the dielectric material, you can treat the new capacitor to be made of two capacitors, one with air as dielectric and the other with the introduced material as the dielectric. But the area of each capacitor is half that of the full capacitor. Further, the two capacitors are in parallel and hence the capacitance (C’) of the new capacitor is given by

C’ = ε₀A/2d + Kε₀A/2d

Putting C = ε₀A/d, we obtain C’ = (K + 1)C/2.

We will discuss more questions on capacitors in due course. Meanwhile, find some useful and interesting multiple choice questions (with solution) at physicsplus.

AP Physics B & C – Electrostatics – Capacitors – Equations to be remembered

The following points are to be noted by AP Physics B as well as AP Physics C aspirants:

(1) Capacitance (C) of a capacitor is given by

C = Q/V where Q is the magnitude of charge on one of the plates of the capacitor and V is the potential difference between the plates. Q is in coulomb, V is in volts and C is in farad.

In a charged capacitor the charge on one plate is positive and the charge on the other plate is negative; but the charges are of equal magnitude so that the total charge on the two plates taken together is zero. But when you say “charge on a capacitor”, you mean the magnitude of charge on one of the plates.

[Often you may be required to calculate the capacitance of a single conductor such as a sphere. What you mean here is the ratio of the charge given to the conductor to the potential to which it is raised: C = Q/V]

(2) Capacitance (C) of a spherical conductor of radius R is given by

C = 4πε₀R

This follows from C = Q/V where V = (1/4πε₀ )(Q/R), which is the potential on the surface of a spherical conductor carrying charge Q.

(3) Capacitance (C) of a parallel plate capacitor having air (or vacuum) as dielectric, with each plate of area A and with separation d between the plates is given by

C = ε₀A/d

If the dielectric is a material of dielectric constant (relative permittivity) K, the capacitance is K times and is given by

C = ε₀ KA/d

If a dielectric slab of dielectric constant K and thickness t is introduced in between the plates of a parallel plate air capacitor, the capacitance (C) is given by

C = ε₀ A/[d – t + (t/K)]

If the space between the plates of a parallel plate capacitor is completely filled with different dielectric slabs of dielectric constants K₁, K₂, K₃, K₄ etc. with thicknesses t₁, t₂, t₃, t₄ etc., the effective capacitance (C) is given by

C = ε₀ A/[d – (t₁ + t₂ + t₃ + t₄ + …) + (t₁ /K₁) +(t₂ /K₂) + (t₃ /K₃) + (t₄ /K₄) +…. ]

Since (t₁ + t₂ + t₃ + t₄ + …) = d, we obtain

C = ε₀ A/[(t₁ /K₁) +(t₂ /K₂) + (t₃ /K₃) + (t₄ /K₄) +…]

(4) Effective capacitance (C) of a series combination of n capacitors of capacitance C₁, C₂, C₃, C₄,….. C_n is given by the reciprocal relation,

1/C = 1/C₁ + 1/C₂ + 1/C₃ + 1/C₄ + ....... + 1/C_n

(5) Effective capacitance (C) of a parallel combination of n capacitors is given by

C = C₁ + C₂ + C₃ + C₄ + ....... + C_n

(6) Energy (U) stored in a charged capacitor is given by

U =(½)CV²

Since Q = CV, the energy can be written also as

U = Q²/2C = (½)QV

Note that the energy in a charged capacitor is stored in the electric field between the plates.

The following points are meant for AP Physics C aspirants only:

(7) If E is the electric field between the plates of a parallel plate capacitor the expression for the energy can be written in terms of the electric field E between the plates as

U = (½)ε₀E²Ad

[You will get it from U = Q²/2C on substituting Q = Aσ and E = σ/ε₀ where σ is the surface density of charge on the plates].

The energy density in the space between the plates is (½)ε₀E² since Ad is the volume of the space between the plates.

(8) Capacitance of a cylindrical capacitor is given by

C = 2πε₀ L / ln(r_b – r_a) where L is the length of the cylinders and r_a and r_b are respectively the radii of the inner and outer cylinders.

(9) A spherical capacitor is made of two concentric spherical conducting shells. Capacitance of a spherical capacitor is given by

C = 4πε₀ r_a r_b /(r_b – r_a) where r_a and r_b are respectively the radii of the inner and outer spherical shells.

In the next post we will discuss questions in this section. Meanwhile, find a useful post containing multiple choice questions (with solution) in this section at physicsplus

AP Physics B and C- Electrostatics: Answers to Free Response Practice Questions on Electric Field and Potential

The world is a dangerous place, not because of those who do evil, but because of those who look on and do nothing.
- Albert Einstein

In the post dated 5^th July 2008, two free response practice questions on electric field and potential were given to you. As promised, I give below model answers along with the questions:

(1) Two identical thin rings of radius R are arranged coaxially with a separation R. A charge +Q is sprayed uniformly on one ring while a charge –Q is sprayed uniformly on the other.

(a) Calculate the electric potentials at the centres of the rings

(b) What is the electric potential on the common axis, midway between the rings? Justify your answer.

(c) A point charge +q is moved very slowly from the centre of the positively charged ring to the cenre of the negatively charged ring. Calculate the work done by the external agency for moving this charge.

(d) The charge +q and the negatively charged ring are now moved to a very large distance from the positively charged ring. Now, determine the electric field at the cntre of the positively charged ring.

(a) The electric potential (V₁) at the centre of the positively charged ring is the sum of the positive potential due to its own positive charge and the negative potential due to the negative charge on the other ring. Thus, putting k = 1/4πε₀ where ε₀ is the permittivity of free space (or air) , we have

V₁ = k[(Q₁/R) + (–Q₂/√2 R)] = k[(Q₁/R) – (Q₂/√2 R)]

[The distance of the charges on the negatively charged ring from the centre of the positively charged ring is R√2].

Similarly, the electric potential (V₂) at the centre of the negatively charged ring is

V₂ = k[(–Q₂/R) + (Q₁/√2 R)] = k[(Q₁/√2 R) – (Q₂/R)]

(b) The electric potential on the common axis, midway between the rings is zero since the positive and negative charges on the rings produce positive and negative potentials of equal magnitude.

(c) The work (W) done in moving the positive charge +q from the centre of one ring to the centre of the other is the product (V₁ – V₂)q.

Therefore, W = k[{(Q₁/R) –(Q₂/√2 R)} – {(Q₁/√2 R) – (Q₂/R)}]q

Or, W = kq(Q₁ + Q₂) [(1/R) – (1//√2 R)] = (kq/√2 R) (Q₁ + Q₂) (√2 – 1)

The work done by the external agency moving the charge is negative since the positive charge will move by itself from the positive potential point to the negative potential point. So, the answer is – W = (kq/√2 R) (Q₁ + Q₂) (1 – √2)

(d) The electric field at the centre of the isolated positively charged ring is zero since a test charge placed at the centre of the ring will feel equal radial forces all around.

(2) Two small identical conducting spheres of radius R carrying charges Q₁ and Q₂ (Q₁ > Q₂) attract each other with a force of magnitude F when they are separated by a distance r (between their centres) in air. When the spheres are brought into contact and then separated to the initial distance r, they repel each other with a force of the same magnitude F. Now, answer the following questions:

(a) It is given that the charge Q₁ is positive. Calculate the electric potential difference between the centres of the spheres before bringing them into contact.

(b) What are the charges on the two spheres after bringing them into contact? Justify your answer.

(c) Calculate the electric field midway between the spheres after bringing the spheres into contact and separating them to the distance r.

(d) Calculate the ratio of the initial charges (Q₁/Q₂) on the spheres.

(e) Keeping the separation at r itself, the charges on the spheres are now changed so that one sphere carries positive charge +4q and the other sphere carries negative charge –q. Calculate the distance of the null point (where the electric field is zero) from the centre of the negatively charged sphere.

(a) The electric potential (V₁) at the centre of the positively charged sphere is the sum of the positive potential due to its own positive charge and the negative potential due to the negative charge on the other sphere. But the potential inside a conductor is the same as the potential on its surface (since there cannot be a field within a conductor). Therefore, the potential at the centre of the positively charged sphere carrying charge Q₁ is given by

V₁ = k[(Q₁/R) –Q₂/(r – R)] where k = 1/4πε₀.

[The sign of Q₂ is negative as indicated by the attractive force between the spheres. The distance of the surface of the positively charged sphere from the centre of the negatively charged sphere is r – R and hence it appears in the above expression for potential].

The electric potential (V₂) at the centre of the negatively charged sphere is given by

V₂ = k[(– Q₂/R) + Q₁/(r – R)]

Potential difference between the cenres of the spheres is

V₁ – V₂ = k[(Q₁/R) –Q₂/(r – R) – {(– Q₂/R) + Q₁/(r – R)}]

= k[Q₁(r – R – R) + Q₂(r – R – R)] /[R(r – R)]

= k (r – 2R) (Q₁ + Q₂) / [R(r – R)]

(b) When the spheres are brought into contact, they share the total charge equally since the spheres are identical. Since the total charge is Q₁– Q₂, the charge on each sphere will be (Q₁– Q₂)/2.

(c) The electric field midway between the spheres is zero since the spheres carry like charges (positive here) of the same magnitude so that a test charge placed at the point will experience equal and opposite forces.

(d) Since the initial attractive force and the final repulsive force between the spheres have the same magnitude F, we have, considering only the magnitudes of the charges,

F = kQ₁Q₂/r² = k[(Q₁– Q₂)/2]²/r² where k =1/4πε₀.

Therefore, Q₁Q₂ = (Q₁– Q₂)²/4.

Or, 4Q₁Q₂ = Q₁²+ Q₂² – 2 Q₁Q₂.

This can be written as Q₁Q₂[(Q₁/ Q₂) + (Q₂/ Q₁) – 6] = 0

Since Q₁Q₂ is not equal to zero, we have

(Q₁/ Q₂) + (Q₂/ Q₁) – 6 = 0

Putting Q₁/ Q₂ = x we have x + 1/x – 6 = 0

Or, x² – 6x + 1 = 0

This gives x = 3 ± √8

The ratio of the magnitudes of the charges (Q₁/ Q₂) is thus 3 ± √8. Since Q₂ is negative, the answer is – (3 ± √8).

So, two values are possible for the ratio of charges.

(e) The null point is nearer the smaller charge, but outside on the straight line passing through the centres of the spheres. The null point is produced because of the equal and opposite forces produced (on a test charge) because of the charges +4q and –q. If the distance of the null point from the centre of the negatively charged sphere is d, the distance from the centre of the positively charged sphere is r+d. Since the net electric field at the null point is zero, we have

k×4q/(r+d)² + k(– q) /d² = 0

Or, 2/(r+d) = 1/d, which gives d = r.

AP Physics B and C- Electrostatics: Free Response Practice Questions on Electric Field and Potential

The following free response practice question which carries 15 points is to be answered in about 17 minutes and is meant for AP Physics B. But it will be useful to AP Physics C aspirants as well:

Two identical thin rings of radius R are arranged coaxially with a separation R. A charge +Q is sprayed uniformly on one ring while a charge –Q is sprayed uniformly on the other.

(a) Calculate the electric potentials at the centres of the rings

(b) What is the electric potential on the common axis, midway between the rings? Justify your answer.

(c) A point charge +q is moved very slowly from the centre of the positively charged ring to the cenre of the negatively charged ring. Calculate the work done by the external agency for moving this charge.

(d) The charge +q and the negatively charged ring are now moved to a very large distance from the positively charged ring. Now, determine the electric field at the cntre of the positively charged ring.

The following free response practice question which carries 15 points is to be answered in about 15 minutes and is meant for AP Physics C aspirants:

Two small identical conducting spheres of radius R carrying charges Q₁ and Q₂ (Q₁ > Q₂) attract each other with a force of magnitude F when they are separated by a distance r (between their centres) in air. When the spheres are brought into contact and then separated to the initial distance r, they repel each other with a force of the same magnitude F. Now, answer the following questions:

(a) It is given that the charge Q₁ is positive. Calculate the electric potential difference between the centres of the spheres before bringing them into contact.

(b) What are the charges on the two spheres after bringing them into contact? Justify your answer.

(c) Calculate the electric field midway between the spheres after bringing the spheres into contact and separating them to the distance r.

(d) Calculate the ratio of the initial charges (Q₁/Q₂) on the spheres.

(e) Keeping the separation at r itself, the charges on the spheres are now changed so that one sphere carries positive charge +4q and the other sphere carries negative charge –q. Calculate the distance of the null point (where the electric field is zero) from the centre of the negatively charged sphere.

Try to answer the above questions. I will be back soon with model answers for your benefit.