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Electric field direction on a grounded conducting sphere
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[QUOTE="zapman345, post: 6314337, member: 584394"] 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. [/QUOTE]
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Electric field direction on a grounded conducting sphere
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