Parallel Plate Capacitor Questions

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Very Short Questions Answers [1 Mark Questions]

Ques. What is meant by a parallel plate capacitor?

Ans. A parallel plate capacitor consists of two parallel conducting plates separated by a small distance connected across a battery.

Ques. The parallel plate capacitor formula is

1. C = kϵ₀d / A
2. C = kϵ₀A / d
3. C = kϵ₀d² / A
4. C = kϵ₀d / A²

Ans. The correct answer is a. C = kϵ₀A / d

Explanation: The capacitance of a parallel plate capacitor when a dielectric medium of dielectric constant k is inserted between the plates of the capacitor is given by C = kϵ₀A/d.

Ques. The two conducting plates of parallel plate capacitors act as electrodes.

1. True
2. False

Ans. The correct answer is a. True

Explanation: Parallel plate capacitors include two conducting plates that act as electrodes with a dielectric between them.

Ques. Units of capacitance are ______

1. Pico-farads (pF)
2. Farad
3. Microfarads (µF)
4. All of the above

Ans. The correct answer is d. All of the above

Explanation: Farad is the SI unit of capacitance. Commonly used units of capacitance are pico farad (pF) and microfarad (µF).

Ques. What are the types of capacitors?

Ans. The different types of capacitors are Ceramic Capacitors, Power Film Capacitors, Ceramic capacitors, Paper Capacitors, Electrolytic Capacitors, Film capacitors, Electrolytic capacitors, and Film Capacitors.

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Short Questions Answers [2 Marks Questions]

Ques. Explain conducting and non-conducting materials with examples.

Ans. The materials that conduct electricity are called conducting materials. A conducting material has a large number of free electrons. Iron, Copper, Silver, etc. are examples of conducting materials.

The materials which do not conduct electricity are called non-conducting materials. A non-conducting material has a very small or negligible number of free electrons. Wood, Plastic, Rubber, etc. are examples of non-conducting materials.

Ques. What is a capacitor?

Ans. A capacitor is a device that stores electrical energy in an electric field by collecting electric charges on two adjacent isolated surfaces. It is a two-terminal passive electrical component.

Ques. Define capacitance. What is the dimensional formula of capacitance?

Ans. The measure of the ability of a capacitor to store electric charge is known as its capacitance. It is also defined as the charge required to raise the potential of a capacitor through one unit.

The dimensional formula of capacitance is [C] = [M^-1 L^-2 T⁴ A²]

Ques. Define dielectric.

Ans. A dielectric is either an insulator or an extremely poor conductor of electric current. When put in an electric field, dielectrics conduct almost zero current because, unlike metals, they contain no loosely bonded, or free, electrons that can drift through the substance.

Also Read:

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Long Questions Answers [3 Marks Questions]

Ques. A parallel plate capacitor is charged to a potential difference of 50 volts. It is then discharged for some time so that its potential drops by 10 volts. Calculate the fraction of energy stored in the capacitor now with respect to the initial energy stored.

Ans. Given

• The initial potential difference between the plates of the capacitor, V_i = 50 volts
• The final potential difference between the plates of the capacitor, V_f = 50 - 10 = 40 volts

Let C be the capacitance of the capacitor, then

Initial potential energy of the capacitor, E_i = CV_i² / 2 = C(50)²/2 = 1250C

Final potential energy of the capacitor, E_f = CV_f² / 2 = C(40)²/2 = 800C

The fraction of energy stored in the capacitor now with respect to the initial energy stored is given by

E_f / E_i = 800C / 1250C = 0.64

Ques. A parallel-plate capacitor with plate area A and separation between the plates d is charged by a constant current i. Consider a plane surface of area A/2 parallel to the plates and drawn symmetrically between the plates. Find the displacement current through this area.

Ans. Given

• Charge on the plate of the capacitor = q
• A is the area of the plates
• The constant current through which the capacitor is charged is i
• d is the separation between the plates

Electric field between the plates, E = q / ϵ₀A

When a plane surface of area A/2 is inserted between the plates, then the net flux through the area is given by

ϕ = E x A/2 = (q / ϵ₀A) x (A/2) = q/2ϵ₀

Now, the displacement current is given by

I_d = ϵ₀ dϕ/dt

⇒ I_d = ϵ₀ d(q/2ϵ₀)/dt

⇒ I_d = 1/2 dq/dt

⇒ I_d = i/2

Hence, the displacement current is i/2.

Ques. If the separation between the plates of a parallel plate capacitor of capacitance 3 F is 5 mm. What is the area of the plates of the capacitor?

Ans. Given

• Capacitance of the parallel plate capacitor, C = 3 F
• The distance of separation between the plates, d = 5 mm = 5 x 10^-3 m

Let A be the area of the plates of the capacitor, then the capacitance of a parallel plate capacitor is given by

C = Aϵ₀/d

⇒ A = C_d/ϵ₀

Where ϵ₀ is the absolute permittivity of the free space = 8.85 x 10^-12 C²N^-1m^-2

On substituting the values, we get

A = (3 x 5 x 10^-3) / (8.85 x 10^-12) = 1.694 x 10⁹ m²

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Very Long Questions Answers [5 Marks Questions]

Ques. Derive the expression for the energy stored in a parallel plate capacitor.

Ans. The process of charging a capacitor is equivalent to that of transferring charge from one plate of the capacitor to another plate. Some work must be done to charge a capacitor. This work is stored as electrostatic potential energy in the capacitor.

Let at any instant, a charge q be on the plate of the capacitor, then the potential difference between the plate is given by

V = q/C

If extra charge dq is transferred to the capacitor, then work done to do so is stored as electric potential energy in the capacitor. i.e.

dU = dW = Vdq = (q/C)dq

The total increase in potential energy in charging the capacitor from q = 0 to q = Q is the total energy stored in the capacitor. Therefore

U = ∫dU = ₀∫^Q\(\frac{q}{c}\)dq = \(\frac{1}{C}\)₀∫^Qqdq = \(\frac{1}{C}\)[\(\frac{q^2}{2}\)]₀^Q = \(\frac{1}{C}\)\(\frac{Q^2}{2}\)

U = \(\frac{Q^2}{2C}\)

Now, we have Q = CV, then

U = \(\frac{1}{2} \frac{C^2V^2}{C}\)

U = \(\frac{CV^2}{C}\)

Also, we have C = Q/V, then

U = \(\frac{1}{2} \frac{Q^2}{Q/V}\)

U = \(\frac{QV}{2}\)

Ques. Derive the expression for the energy density in a parallel plate capacitor.

Ans. The energy stored in a parallel plate capacitor is given by

U = CV² / 2 …(i)

But the capacitance of a parallel plate capacitor is given by

C = ϵ₀A / d

Also, the potential difference between the plates of the capacitor is given by

V = Ed

Where

• E is the electric field between the plates of the capacitor
• d is the distance between the plates

Substituting these values in equation (i), we get

U = (ϵ₀A / d)(Ed)² / 2

⇒ U = (1/2)ϵ₀E²Ad

⇒ U / Ad = (1/2)ϵ₀E²

But volume of the capacitor = Ad

⇒ U / volume = (1/2)ϵ₀E²

Energy stored per unit volume of the parallel plate capacitor is known as energy density (U_d).

⇒ U_d = ϵ₀E² / 2

Ques. A fully charged parallel plate capacitor is connected across an uncharged identical capacitor. Show that the energy stored in the combination is less than that stored initially in the single capacitor.

Ans. Let the amount of charge on the charged parallel plate capacitor of capacitance C be Q.

Initially, the energy stored is given by

U_i = Q² / 2C

When an uncharged identical capacitor of capacitance C is connected across the charged capacitor, due to the sharing of charges, the charge of each capacitor becomes Q/2.

The final energy of the combination is given by

U_f = (Q/2)² / 2C + (Q/2)² / 2C

⇒ U_f = Q² / 8C + Q² / 8C = Q² / 4C

It is observed that U_f < U_i therefore the energy stored in the combination is less than that stored initially in the single capacitor.

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