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## Homework Statement

I'm quite lost here.

The problem states:

A spherically symmetric electrostatic charge distribution produces an electric field that is directed radially out from the distribution center. At the distance

*r*from this center the field has the strength:

[tex]\[E=\frac{Q}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\left( \frac{r}{R}-\frac{{{r}^{2}}}{{{R}^{2}}} \right),0<r<R\][/tex]

and

[tex]E=\frac{Q}{4\pi {{\varepsilon }_{0}}{{r}^{2}}},r>R[/tex]

The constant

*Q*is a charge, the constant

*R*is a length; both are positive.

Now, determine the electric potential

*V*(

*r*) anywhere in space, when the potential is set to 0 in the infinitely distant.

## Homework Equations

[tex]\[V=\int_{a}^{b}{E}\,dr\][/tex]

## The Attempt at a Solution

Ok...

I first do the integral with the first

*E*-field (0 <

*r*<

*R*), with the limits 0 to

*r*. This gives me:

[tex]V=\frac{Q}{4\pi {{\varepsilon }_{0}}{{R}^{2}}}\int_{0}^{r}{\left( \frac{r}{R}-\frac{{{r}^{2}}}{{{R}^{2}}} \right)}\,dr=-\frac{Q{{r}^{2}}\left( 2r-3R \right)}{24\pi {{\varepsilon }_{0}}{{R}^{4}}}[/tex]

But since

*Q*and

*R*were positive, I assume that this result is positive as well, so instead we have:

[tex]V=\frac{Q{{r}^{2}}\left( 2r-3R \right)}{24\pi {{\varepsilon }_{0}}{{R}^{4}}}[/tex]

Then I take the integral of the other E-field, with the limits

*R*to infinity and get:

[tex]V=\frac{Q}{4\pi {{\varepsilon }_{0}}}\int_{R}^{\infty }{\frac{1}{{{r}^{2}}}}\,dr=\frac{Q}{4\pi {{\varepsilon }_{0}}R}[/tex]

Adding these two integrals together I get:

[tex]V=\[\frac{Q\left( 2{{r}^{3}}-3R{{r}^{2}}+6{{R}^{3}} \right)}{24\pi {{\varepsilon }_{0}}{{R}^{4}}}\][/tex]

BUT, according to my book it should be:

[tex]V=\[\frac{Q\left( 2{{r}^{3}}-3R{{r}^{2}}+7{{R}^{3}} \right)}{24\pi {{\varepsilon }_{0}}{{R}^{4}}}\][/tex]

So, the difference being the 7 instead of 6. And I have no idea what to do to make up for that tiny difference. It's so close to each other that I can't imagine me being WAY off. So am I doing something wrong, or is it a mistake in the book ?

Regards