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That's a hollow conducting sphere link

My charge is at P (10uC). R = 0.15m. PS = 0.05m. DS = 0.35m. What's the electric field at point D? enter image description here

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  • $\begingroup$ Is that the only charge in the sphere? $\endgroup$ Mar 23, 2013 at 18:26
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Assuming that the point charge is the only charge in the sphere, the electric field will simply follow from Gauss's law:

$$\vec{E}=\frac{1}{4\pi\epsilon}\frac{q_p}{r_{PD}^2}\hat{r}_{PD}$$

Where $\hat{r}_{PD}$ is a unit vector pointing from P to D. Note that in this equation, $\epsilon$ is not the permittivity of free space, but rather the permittivity of the material in the sphere. So there will be a dielectric constant $k$ associated with the sphere, and the permittivity can be calculated as $k\epsilon_0$, where $\epsilon_0$ is $8.85 \times10^{-12}\frac{F}{m}$, as usual.

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