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From Griffiths, Third edition Intro Electrodynamics

I've been working on this problem, and I get 2 different answers. I get the same answer for parts b and c, but not for a. I know I'm probably just doing something silly along the way, but I can't find my mistake...

Find the energy stored in a uniformly charged solid sphere of radius R and charge q. Do it three different ways:

(a) Use Eq. 2.43

(b) Use Eq. 2.45

(c) Use Eq. 2.44

Eq. 2.43 [tex]W=\frac{1}{2}\int \rho V d\tau[/tex]

Eq. 2.45 [tex]W=\frac{\epsilon_{0}}{2}\int_{all space} E^2 d\tau[/tex]

Eq. 2.44 [tex]W=\frac{\epsilon_{0}}{2}\left(\int_{V} E^2 d\tau + \oint_{S} VE\cdot da\right)[/tex]

For part (a) we have:

[tex] E(r)=\left\{\begin{array}{cc}\frac{kq}{r^2}, &\mbox{ if } r > R\\ \frac{kq}{R^3} r, &\mbox{ if } r < R \end{array}\right

\\ \Rightarrow V(r) = \left\{\begin{array}{cc}\frac{kq}{r}, & r > R\\ \frac{3kq}{2R} - \frac{kq}{R^3} r^2, & r < R \end{array}\right[/tex]

Since it's uniformly charged, we know that:

[tex] \rho = \frac{3q}{4\pi R^3} \\ [/tex]

Then we can evaluate Eq. 2.43 to get:

[tex] W=2\pi\rho\int_{0}^R V(r) r^2 dr \\

= \frac{9}{20} \frac{kq^2}{R} [/tex]

For (b) and (c) however I get:

[tex] W=\frac{3}{5} \frac{kq^2}{R} [/tex]

So they are off by a factor of 3/4.

If you need to see more work to help me find my mistake, let me know and I will post more details.

I've been working on this problem, and I get 2 different answers. I get the same answer for parts b and c, but not for a. I know I'm probably just doing something silly along the way, but I can't find my mistake...

## Homework Statement

Find the energy stored in a uniformly charged solid sphere of radius R and charge q. Do it three different ways:

(a) Use Eq. 2.43

(b) Use Eq. 2.45

(c) Use Eq. 2.44

## Homework Equations

Eq. 2.43 [tex]W=\frac{1}{2}\int \rho V d\tau[/tex]

Eq. 2.45 [tex]W=\frac{\epsilon_{0}}{2}\int_{all space} E^2 d\tau[/tex]

Eq. 2.44 [tex]W=\frac{\epsilon_{0}}{2}\left(\int_{V} E^2 d\tau + \oint_{S} VE\cdot da\right)[/tex]

## The Attempt at a Solution

For part (a) we have:

[tex] E(r)=\left\{\begin{array}{cc}\frac{kq}{r^2}, &\mbox{ if } r > R\\ \frac{kq}{R^3} r, &\mbox{ if } r < R \end{array}\right

\\ \Rightarrow V(r) = \left\{\begin{array}{cc}\frac{kq}{r}, & r > R\\ \frac{3kq}{2R} - \frac{kq}{R^3} r^2, & r < R \end{array}\right[/tex]

Since it's uniformly charged, we know that:

[tex] \rho = \frac{3q}{4\pi R^3} \\ [/tex]

Then we can evaluate Eq. 2.43 to get:

[tex] W=2\pi\rho\int_{0}^R V(r) r^2 dr \\

= \frac{9}{20} \frac{kq^2}{R} [/tex]

For (b) and (c) however I get:

[tex] W=\frac{3}{5} \frac{kq^2}{R} [/tex]

So they are off by a factor of 3/4.

If you need to see more work to help me find my mistake, let me know and I will post more details.

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