krit.club logo

Electrostatic Potential and Capacitance - Capacitors and Capacitance

Grade 12CBSEPhysics

Review the key concepts, formulae, and examples before starting your quiz.

πŸ”‘Concepts

β€’

A capacitor is a system of two conductors separated by an insulator, used to store electrical charge and electrical potential energy.

β€’

Capacitance (CC) is defined as the ratio of the magnitude of charge (QQ) on either conductor to the potential difference (VV) between them: C=QVC = \frac{Q}{V}. The SI unit is Farad (FF).

β€’

The capacitance of a parallel plate capacitor depends on the area (AA) of the plates and the distance (dd) between them. It is given by C=Ο΅0AdC = \frac{\epsilon_0 A}{d} in vacuum.

β€’

When a dielectric medium of dielectric constant KK is completely filled between the plates, the capacitance increases by a factor of KK, becoming C=KΟ΅0AdC = \frac{K \epsilon_0 A}{d}.

β€’

In a series combination of capacitors, the reciprocal of the equivalent capacitance is the sum of the reciprocals of individual capacitances. The charge QQ remains the same across each capacitor.

β€’

In a parallel combination of capacitors, the equivalent capacitance is the sum of the individual capacitances. The potential difference VV remains the same across each capacitor.

β€’

The energy stored in a capacitor is the work done in charging it, which is stored as electrostatic potential energy UU in the electric field between the plates.

β€’

Energy density (uu) is the energy stored per unit volume in the electric field, expressed as u=12Ο΅0E2u = \frac{1}{2} \epsilon_0 E^2.

πŸ“Formulae

C=QVC = \frac{Q}{V}

C=Ο΅0AdC = \frac{\epsilon_0 A}{d}

Cmedium=KCvacuumC_{medium} = K C_{vacuum}

1Cs=1C1+1C2+1C3+…\frac{1}{C_s} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} + \dots

Cp=C1+C2+C3+…C_p = C_1 + C_2 + C_3 + \dots

U=12CV2=Q22C=12QVU = \frac{1}{2}CV^2 = \frac{Q^2}{2C} = \frac{1}{2}QV

u=12Ο΅0E2u = \frac{1}{2} \epsilon_0 E^2

πŸ’‘Examples

Problem 1:

A parallel plate capacitor with air between the plates has a capacitance of 8Β pF8 \text{ pF}. What will be the capacitance if the distance between the plates is reduced by half, and the space between them is filled with a substance of dielectric constant K=6K = 6?

Solution:

Original capacitance C0=Ο΅0Ad=8Β pFC_0 = \frac{\epsilon_0 A}{d} = 8 \text{ pF}. New distance dβ€²=d2d' = \frac{d}{2} and dielectric constant K=6K = 6. The new capacitance Cβ€²=KΟ΅0Adβ€²=6Ο΅0Ad/2=12(Ο΅0Ad)=12Γ—8Β pF=96Β pFC' = \frac{K \epsilon_0 A}{d'} = \frac{6 \epsilon_0 A}{d/2} = 12 \left( \frac{\epsilon_0 A}{d} \right) = 12 \times 8 \text{ pF} = 96 \text{ pF}.

Explanation:

Capacitance is directly proportional to the dielectric constant and inversely proportional to the distance between plates.

Problem 2:

Three capacitors of capacitances 2Β ΞΌF2 \text{ } \mu F, 3Β ΞΌF3 \text{ } \mu F, and 4Β ΞΌF4 \text{ } \mu F are connected in parallel. (a) What is the total capacitance? (b) Determine the charge on each capacitor if the combination is connected to a 100Β V100 \text{ V} supply.

Solution:

(a) For parallel combination: Ceq=C1+C2+C3=2+3+4=9Β ΞΌFC_{eq} = C_1 + C_2 + C_3 = 2 + 3 + 4 = 9 \text{ } \mu F. (b) In parallel, VV is same for all. Q1=C1V=2Γ—10βˆ’6Γ—100=2Γ—10βˆ’4Β CQ_1 = C_1 V = 2 \times 10^{-6} \times 100 = 2 \times 10^{-4} \text{ C}, Q2=C2V=3Γ—10βˆ’6Γ—100=3Γ—10βˆ’4Β CQ_2 = C_2 V = 3 \times 10^{-6} \times 100 = 3 \times 10^{-4} \text{ C}, Q3=C3V=4Γ—10βˆ’6Γ—100=4Γ—10βˆ’4Β CQ_3 = C_3 V = 4 \times 10^{-6} \times 100 = 4 \times 10^{-4} \text{ C}.

Explanation:

In parallel circuits, the voltage across each capacitor is equal to the supply voltage, and the total capacitance is the simple sum.

Problem 3:

A 12Β pF12 \text{ pF} capacitor is connected to a 50Β V50 \text{ V} battery. How much electrostatic energy is stored in the capacitor?

Solution:

Given C=12Β pF=12Γ—10βˆ’12Β FC = 12 \text{ pF} = 12 \times 10^{-12} \text{ F} and V=50Β VV = 50 \text{ V}. Energy U=12CV2=12Γ—(12Γ—10βˆ’12)Γ—(50)2=6Γ—10βˆ’12Γ—2500=1.5Γ—10βˆ’8Β JU = \frac{1}{2} CV^2 = \frac{1}{2} \times (12 \times 10^{-12}) \times (50)^2 = 6 \times 10^{-12} \times 2500 = 1.5 \times 10^{-8} \text{ J}.

Explanation:

The formula U=12CV2U = \frac{1}{2} CV^2 is used to calculate the energy stored in the electric field of the capacitor.

Capacitors and Capacitance Class 12 Notes & Examples