Electric field direction on a grounded conducting sphere

[zapman345]

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

 

[haruspex]

Don't you need to evaluate $\frac{\partial V}{\partial\theta}$ at r=R?

 

[zapman345]

I did but that wouldn't make the $\theta$ component vanish cause I find that $$E_{\theta} = -\frac{1}{r}\cdot\left(\frac{rR^{3}\sin(\theta)}{d^{2}\left(\frac{R^{4}}{d^{2}}-\frac{2rR^{2}\cos(\theta)}{d}+r^{2}\right)^{3/2}}-\frac{dr\sin(\theta)}{\left(d^{2}-2dr\cos(\theta)+r^{2}\right)^{3/2}}\right)\hat{\theta}$$ Which even at $r=R$ doesn't vanish, it does vanish for $\theta=0$ but that is a trivial solution as that is the line directly between the real and fictive charge used in the method of images. For any $\theta\neq0$ this won't vanish but intuitively I think it should vanish because we're talking about a spherical object.

 

[hutchphd]

Consider the case R →∞ , you should recover the flat plane case for the potential. I think you need to worry about the angle for the image charge...you seem have a sign wrong.

 

[zapman345]

You were right. The potential I had calculated should be $$
\begin{align*}

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

\end{align*}$$
Then for the electric field you will find $$\begin{align*}
E & =-\frac{q}{4\pi\epsilon_{0}}\left\{ \left(\frac{2\left(\frac{d}{R}\right)^{2}r-2d\cos(\theta)}{2\left(R^{2}+\left(\frac{dr}{R}\right)^{2}-2ar\cos(\theta)\right)^{3/2}}+\frac{2r-2d\cos(\theta)}{2\left(r^{2}+d^{2}-2dr\cos(\theta)\right)^{3/2}}\right)\hat{r}+\frac{1}{r}\cdot\left(\frac{dr\sin\left(\theta\right)}{\left(R^{2}+\left(\frac{rd}{R}\right)^{2}-2dr\cos\left(\theta\right)\right)^{3/2}}-\frac{dr\sin\left(\theta\right)}{\left(r^{2}+d^{2}-2dr\cos\left(\theta\right)\right)^{3/2}}\right)\hat{\theta}\right\}

\end{align*}$$

Which reduces to
\begin{align*}

E\left(r=R\right) & =-\frac{q}{4\pi\epsilon_{0}}\left(\frac{d^{2}-R^{2}}{R\left(R^{2}+d^{2}-2dR\cos(\theta)\right)^{3/2}}\hat{r}+0\hat{\theta}\right)\\

& =-\frac{q}{4\pi\epsilon_{0}}\left(\frac{d^{2}-R^{2}}{R\left(R^{2}+d^{2}-2dR\cos(\theta)\right)^{3/2}}\right)\hat{r}

\end{align*}

Pointing only in the radial direction on the surface as expected. Thank you for all your help.