Question 840: Electrostats & Capacitance

Question

The electric field intensity on the surface of a solid charged sphere of radius '$$r$$' and volume charge density '$$\rho$$' is ($$\varepsilon_0=$$ permittivity of free space)

Accepted Answer

According to Gauss' theorem,
$$\phi=\mathrm{E} \cdot \mathrm{A}=\frac{\mathrm{q}_{\mathrm{enc}}}{\varepsilon_0}$$
But, $$\mathrm{q}_{\mathrm{enc}}=ho 	imes$$ volume
$$	herefore \quad \mathrm{q}_{\mathrm{enc}}=ho\left(\frac{4}{3} \pi \mathrm{r}^3ight)$$
And area is $$A=4 \pi r^2$$
$$\begin{aligned}
& 	herefore \quad E\left(4 \pi r^2ight)=\frac{ho\left(\frac{4}{3} \pi r^3ight)}{\varepsilon_0} \
& 	herefore \quad E=\frac{ho r}{3 \varepsilon_0}
\end{aligned}$$

Answer

$$\frac{\rho r}{3 \varepsilon_0}$$

Answer

$$\frac{\rho}{4 \pi \varepsilon_0 \mathrm{r}}$$

Answer

zero

Answer

$$\frac{5 \rho \mathrm{r}}{6 \varepsilon_0}$$

Electrostats & Capacitance

The electric field intensity on the surface of a solid charged sphere of radius '$r' and volume charge density '\rho' is (\varepsilon_0=$ permittivity of free space)

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