Capacitance and Dielectrics — Part 2: Energy and dielectrics: visual lesson, ISC Class 12 Physics

Lesson

Energy and dielectrics

Part 2 — how much energy a capacitor stores, and how an insulating slab between the plates increases its capacitance.

In this lesson you will

Calculate the energy stored in a capacitor and the energy density of the field
Explain polarisation and predict the effect of a dielectric with the battery connected or disconnected
Calculate the capacitance of a capacitor partly filled with a dielectric slab

Concept

Energy stored in a capacitor

Charging a capacitor needs work against the growing potential difference; this work is stored as electric potential energy: $U = \dfrac{1}{2}CV^2 = \dfrac{1}{2}QV = \dfrac{Q^2}{2C}$

On the V–Q graph (a straight line), U is the area under the line.
The energy is stored in the field between the plates: energy density $u = \dfrac{1}{2}\varepsilon_0E^2$ (J m⁻³).

V = Q/C (C = 2.0 μF)

Area under the V–Q line = work done in charging = ½QV.

Worked example

Worked example 4: energy stored and energy density

(a) Find the energy stored in a 5.0 μF capacitor charged to 200 V. (b) Find the energy density in a field of $2.0 \times 10^{5}$ V m $^{-1}$ in air.

(a) $U = \tfrac12CV^2 = 0.5 \times 5.0 \times 10^{-6} \times (200)^2 = 0.10$ J.
(b) $u = \tfrac12\varepsilon_0E^2 = 0.5 \times 8.85 \times 10^{-12} \times (2.0 \times 10^{5})^2 \approx 0.177$ J m $^{-3}$ .

V = Q/C (C = 2.0 μF)

Area under the V–Q line = work done in charging = ½QV.

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