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## Homework Statement:

- Is the angle that the electric field makes on the surface of a grounded conducting sphere near a (positive) point charge 90°? I.e. is the electric field pointed radially outward from the sphere.

## Relevant Equations:

- ##E=-\nabla V##

I am required to find the direction of the electric field on the surface of a grounded conducting sphere in the proximity of a point charge ##+q##. The distance between the center of the sphere and the point charge is ##d## and using the method of images we find that the charge of the sphere is ##q' = -\frac{R}{d}q## then we can calculate the potential which I found here:

$$V\left(\vec{r}\right) =\frac{q}{4\pi\epsilon_{0}}\left[\frac{1}{\sqrt{d^{2}+r^{2}-2dr\cos\left(\theta\right)}}-\frac{\frac{R}{d}}{\sqrt{\left(\frac{R^{2}}{d}\right)^{2}+r^{2}-2\left(\frac{R^{2}}{d}\right)r\cos\left(\theta\right)}}\right]$$

Now we also know that the electric field is:

\begin{align*}

E & =-\nabla V\\

& =-\left(\frac{\partial V}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial V}{\partial\theta}\hat{\theta}+\frac{1}{r\sin\left(\theta\right)}\frac{\partial V}{\partial\phi}\hat{\phi}\right)\\

& =-\left(\frac{\partial V}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial V}{\partial\theta}\hat{\theta}\right)\\

\end{align*}

Can I from this positively say that the electric field outside the sphere is

$$V\left(\vec{r}\right) =\frac{q}{4\pi\epsilon_{0}}\left[\frac{1}{\sqrt{d^{2}+r^{2}-2dr\cos\left(\theta\right)}}-\frac{\frac{R}{d}}{\sqrt{\left(\frac{R^{2}}{d}\right)^{2}+r^{2}-2\left(\frac{R^{2}}{d}\right)r\cos\left(\theta\right)}}\right]$$

Now we also know that the electric field is:

\begin{align*}

E & =-\nabla V\\

& =-\left(\frac{\partial V}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial V}{\partial\theta}\hat{\theta}+\frac{1}{r\sin\left(\theta\right)}\frac{\partial V}{\partial\phi}\hat{\phi}\right)\\

& =-\left(\frac{\partial V}{\partial r}\hat{r}+\frac{1}{r}\frac{\partial V}{\partial\theta}\hat{\theta}\right)\\

\end{align*}

Can I from this positively say that the electric field outside the sphere is

*not*radially pointed outward only, and thus does not make a 90° angle with the surface of the sphere?