Electric Potential Due to A Uniformly Charged Sphere

Hello, I'm trying to find the electric potential due to a uniformly charged sphere, radius R. I use the definition,

V=-\int\limits_{\infty}\limits^{r} \mathbf{E}\cdot\mathrm{d} \boldsymbol{\ell}=+\int\limits_{\infty}\limits^{r}\dfrac{Q}{4 \pi \varepsilon_0 r^2}\ \mathrm{d}r,

because dl is in the -r direction. This gives the negative of the answer I should have, why is this?

(edited 11 years ago)

Reply 1

dknt

Original post by KeyFingot

Hello, I'm trying to find the electric potential due to a uniformly charged sphere, radius R. I use the definition,

V=-\int\limits_{\infty}\limits^{r} \mathbf{E}\cdot\mathrm{d} \boldsymbol{\ell}=+\int\limits_{\infty}\limits^{r}\dfrac{Q}{4 \pi \varepsilon_0 r^2}\ \mathrm{d}r,

because dl is in the -r direction. This gives the negative of the answer I should have, why is this?

You need to also change your limits to correspond to r which would be from r to infinity.

Hopefully you can see this if you write,

V= -\int\limits_{\infty}\limits^{r} \mathbf{E}\cdot\mathrm{d} \boldsymbol{\ell}= + \int\limits_{r}\limits^{\infty} \mathbf{E}\cdot\mathrm{d} \boldsymbol{\ell}

, just by inverting the limits.

Now, from this we can see that the vector dl now goes in the same direction as dr, so we get

\mathrm{d}\boldsymbol{\ell} = +\mathrm{d}\mathbf{r}

V= -\int\limits_{\infty}\limits^{r} \mathbf{E}\cdot\mathrm{d} \boldsymbol{\ell}= + \int\limits_{r}\limits^{\infty} \mathbf{E}\cdot\mathrm{d} \boldsymbol{\ell} = + \int\limits_{r}\limits^{\infty} \mathbf{E}\cdot\mathrm{d} \mathbf{r}

Reply 2

KeyFingot

Original post by dknt

You need to also change your limits to correspond to r which would be from r to infinity.

Hopefully you can see this if you write,

V= -\int\limits_{\infty}\limits^{r} \mathbf{E}\cdot\mathrm{d} \boldsymbol{\ell}= + \int\limits_{r}\limits^{\infty} \mathbf{E}\cdot\mathrm{d} \boldsymbol{\ell}

, just by inverting the limits.

Now, from this we can see that the vector dl now goes in the same direction as dr, so we get

\mathrm{d}\boldsymbol{\ell} = +\mathrm{d}\mathbf{r}

V= -\int\limits_{\infty}\limits^{r} \mathbf{E}\cdot\mathrm{d} \boldsymbol{\ell}= + \int\limits_{r}\limits^{\infty} \mathbf{E}\cdot\mathrm{d} \boldsymbol{\ell} = + \int\limits_{r}\limits^{\infty} \mathbf{E}\cdot\mathrm{d} \mathbf{r}

OK I think I understand but why did my original method not work?

Reply 3

Stonebridge

Original post by KeyFingot

Hello, I'm trying to find the electric potential due to a uniformly charged sphere, radius R. I use the definition,

V=-\int\limits_{\infty}\limits^{r} \mathbf{E}\cdot\mathrm{d} \boldsymbol{\ell}=+\int\limits_{\infty}\limits^{r}\dfrac{Q}{4 \pi \varepsilon_0 r^2}\ \mathrm{d}r,

because dl is in the -r direction. This gives the negative of the answer I should have, why is this?

The integral is correct (second expression in the line) It's from infinity to R as it represents the work done moving a positive charge from infinity to R.
So Q is positive. The work done by the force is F (-dr) where dr is in the negative r direction due to the fact that the force is repulsive and the applied force by the external agent is towards the sphere.
So the negative sign is there in the integral expression. For some reason it has disappeared in your third expression. Why is that?
When you then integrate the negative expression within those limits the expression for electric potential (with 1/r in it) will be positive. This is correct as the potential due to a positively charged sphere is positive at its surface.

Reply 4

KeyFingot

Original post by Stonebridge

The negative sign disappeared because I was integrating from infinity to r, so

\mathrm{d}\boldsymbol{\ell}=-\mathrm{d}r\ \!\hat{r}

which cancels with the minus outside the integral.

Reply 5

KeyFingot

Oh I understand now, I changed the sign on the "dl" part but not the limits hence my answer had a factor of -1 when it shouldn't. Thank you both for your help!