Question 891: Electrostats & Capacitance

Question

Surface density of charge on a charged conducting sphere of radius $$R$$ in terms of electric field intensity $$E$$ at a distance $$r$$ in free space is ($$r>R, \varepsilon_0=$$ permittivity of free space)

Accepted Answer

The electric field intensity at a distance $$r$$ from the conducting sphere is
$$E=\frac{1}{4 \pi \varepsilon_0} \frac{Q}{r^2} \quad 	ext{..... (i)}$$
The surface charge density of sphere is
$$\sigma=\frac{Q}{4 \pi R^2} \quad 	ext{..... (ii)}$$
Dividing Eqs. (ii) by (i), we get
$$\begin{aligned}
& \frac{\sigma}{E}=\frac{Q}{4 \pi R^2} 	imes \frac{4 \pi \varepsilon_0 r^2}{Q}=\varepsilon_0\left(\frac{r}{R}ight)^2 \
	ext{or} \quad & \sigma=\varepsilon_0 E\left(\frac{r}{R}ight)^2
\end{aligned}$$

Answer

$$\varepsilon_0 E\left(\frac{R}{r}\right)^2$$

Answer

$$\varepsilon_0 E \frac{R}{r}$$

Answer

$$\varepsilon_0 E\left(\frac{r}{R}\right)^2$$

Answer

$$\varepsilon_0 E \frac{r}{R}$$

Electrostats & Capacitance

Surface density of charge on a charged conducting sphere of radius $R in terms of electric field intensity E at a distance r in free space is (r>R, \varepsilon_0=$ permittivity of free space)

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