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Electrostatic Potential and Capacitance - Energy Stored in a Capacitor

Grade 12CBSEPhysics

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

🔑Concepts

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The process of charging a capacitor involves transferring electric charges from one plate to another. This requires work to be done against the existing electrostatic force of the charges already present.

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The work done in the charging process is stored as electrostatic potential energy (UU) in the electric field between the plates of the capacitor.

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Energy density (uu) is defined as the energy stored per unit volume of the space between the capacitor plates. For a parallel plate capacitor, it is proportional to the square of the electric field intensity (E2E^2).

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When a dielectric slab of dielectric constant KK is inserted with the battery disconnected, the energy stored decreases to U′=UKU' = \frac{U}{K}.

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If the dielectric slab is inserted while the battery remains connected, the energy stored increases to U′=KUU' = K U because the potential difference remains constant while capacitance increases.

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When two capacitors are connected in parallel, there is a loss of energy (ΔU\Delta U) in the form of heat or electromagnetic radiation due to the redistribution of charges, until they reach a common potential.

📐Formulae

U=12Q2CU = \frac{1}{2} \frac{Q^2}{C}

U=12CV2U = \frac{1}{2} CV^2

U=12QVU = \frac{1}{2} QV

u=12ϵ0E2u = \frac{1}{2} \epsilon_0 E^2

ΔU=C1C2(V1−V2)22(C1+C2)\Delta U = \frac{C_1 C_2 (V_1 - V_2)^2}{2(C_1 + C_2)}

💡Examples

Problem 1:

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

Solution:

Given: C=12 pF=12×10−12 FC = 12 \text{ pF} = 12 \times 10^{-12} \text{ F} and V=50 VV = 50 \text{ V}. Using the formula U=12CV2U = \frac{1}{2} CV^2: U=12×(12×10−12)×(50)2U = \frac{1}{2} \times (12 \times 10^{-12}) \times (50)^2 U=6×10−12×2500U = 6 \times 10^{-12} \times 2500 U=1.5×10−8 JU = 1.5 \times 10^{-8} \text{ J}

Explanation:

The energy is calculated by substituting the capacitance and potential difference into the energy formula. The result represents the energy stored in the electric field between the plates.

Problem 2:

A 600 pF600 \text{ pF} capacitor is charged by a 200 V200 \text{ V} supply. It is then disconnected from the supply and is connected to another uncharged 600 pF600 \text{ pF} capacitor. How much electrostatic energy is lost in the process?

Solution:

Initial energy Ui=12C1V12=12×600×10−12×(200)2=1.2×10−5 JU_i = \frac{1}{2} C_1 V_1^2 = \frac{1}{2} \times 600 \times 10^{-12} \times (200)^2 = 1.2 \times 10^{-5} \text{ J}. When connected to an uncharged capacitor (C2=600 pF,V2=0C_2 = 600 \text{ pF}, V_2 = 0), the common potential V=C1V1+C2V2C1+C2=600×200+01200=100 VV = \frac{C_1 V_1 + C_2 V_2}{C_1 + C_2} = \frac{600 \times 200 + 0}{1200} = 100 \text{ V}. Final energy Uf=12(C1+C2)V2=12×1200×10−12×(100)2=0.6×10−5 JU_f = \frac{1}{2} (C_1 + C_2) V^2 = \frac{1}{2} \times 1200 \times 10^{-12} \times (100)^2 = 0.6 \times 10^{-5} \text{ J}. Energy loss ΔU=Ui−Uf=0.6×10−5 J\Delta U = U_i - U_f = 0.6 \times 10^{-5} \text{ J}.

Explanation:

Energy loss occurs because work is done in moving charges through the connecting wires, which dissipate energy as heat until both capacitors reach the same potential.