## Related questions with answers

A $1-\mu \mathrm{C}$ point charge is embedded in the center of a solid Pyrex sphere of radius $R=10 \mathrm{~cm}$. (a) Calculate the electric field strength $E$ just beneath the surface of the sphere. (b) Assuming that there are no other free charges, calculate the strength of the electric field just outside the surface of the sphere. (c) What is the induced surface charge density $\sigma_{\text {ind }}$ on the surface of the Pyrex sphere?

Solution

Verifieda)

From equation (29-23) we can find the electric field inside the $\textit{Pyrex}$ sphere caused by the point charge:

$\begin{align*} E &= \frac{1}{\kappa_{\text{e}}} E_{0} \\ E &= \frac{1}{\kappa_{\text{e}}} \frac{1}{4 \pi \epsilon_{0}} \frac{q}{R^{2}}\\ E &= \frac{1}{4.7} \frac{1}{4 \pi \epsilon_{0}} \frac{10^{-6} \ \text{C}}{(0.1 \ \text{m})^{2}} \end{align*}$

$\boxed{E \approx 1.91 \cdot 10^{5} \ \frac{\text{V}}{\text{m}}}$

b)

Outside the sphere there is no dielectric reaction on the field and the field is simply $E_{0}$. We can also imagine this on another way, when the dipoles rearrange themselves the total charge on the sphere is equal to the point charge in the middle of the sphere. This gives us the total electric field outside the sphere as it there was no sphere at all:

$\begin{align*} E &= E_{0}\\ E &= \frac{1}{4 \pi \epsilon_{0}} \frac{q}{R^{2}}\\ E &= \frac{1}{4 \pi \epsilon_{0}} \frac{10^{-6} \ \text{C}}{(0.1 \ \text{m})^{2}} \end{align*}$

$\boxed{E \approx 8.99 \cdot 10^{5} \ \frac{\text{V}}{\text{m}}}$

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