A sphere of radius \(R\) carries total charge \(Q\) distributed uniformly over its surface. Show that the energy stored in its electric field is \(U=k Q^{2} / 2 R\).

Short Answer

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The energy stored in the electric field of a sphere of radius R carrying total charge Q distributed uniformly over its surface is given by \(U = k Q^{2} / 2R\)

Step by step solution

01

Understand the basics

The energy stored in an electric field is given by \(E = \frac{1}{2} \int E^2 dv\), where E is the electric field and dv is an infinitesimal volume element over which we integrate. For a surface charge, \(E = k \cdot \frac{Q}{r^2}\), where k is a coulomb's constant, Q is the charge and r is the distance from the center of sphere.
02

Setup the energy integration

Notice that the energy will be different for the region inside and outside the sphere. So, when we setup the integral for energy, we will have two integrals, one from 0 to R and other from R to infinity. The energy is then given by the sum of these two integrals. \(E = \frac{1}{2} \int_{0}^{R} k^2 \cdot \frac{Q^2}{r^4} \cdot 4\pi r^2 dr + \frac{1}{2} \int_{R}^{\infty} k^2 \cdot \frac{Q^2}{r^4} \cdot 4\pi r^2 dr\) Note that we multiplied by \(4\pi r^2\) because that is the surface area element for a spherical shell at radius r.
03

Solve the integrals

Solving these integrals, we get: \(\frac{1}{2} \int_{0}^{R} k^2 \cdot \frac{Q^2}{r^2} dr + \frac{1}{2} \int_{R}^{\infty} k^2 \cdot \frac{Q^2}{r^2} dr\) which simplifies to \(kQ^2 (1 - 1/R)\) for the first and \(kQ^2/R\) for the second.
04

Combine the results

Adding these results, we get the total energy \(U = kQ^2 (1 - 1/R) + kQ^2/R = k Q^{2}/ 2R\). So, the energy stored in its electric field is \(U = k Q^{2} / 2R\).

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