- Home
- Standard 12
- Physics
Let a total charge $2Q$ be distributed in a sphere of radius $R$, with the charge density given by $\rho(r) = kr$, where $r$ is the distance from the centre. Two charges $A$ and $B$, of $-Q$ each, are placed on diametrically opposite points, at equal distance, $a$, from the centre. If $A$ and $B$ do not experience any force, then
$a = \frac{{3R}}{{{2^{1/4}}}}$
$a = {2^{ - 1/4}}R$
$a = {8^{ - 1/4}}R$
$a = R/\sqrt 3 $
Solution
$E 4 \pi a^{2}=\frac{\int_{0}^{\theta} k r 4 \pi r^{2} d r}{e_{0}}$
$E=\frac{k 4 \pi a^{4}}{4 \times 4 \pi \varepsilon_{0}}$
$2 \mathrm{Q}=\int_{0}^{\mathrm{R}} \mathrm{kr} 4 \pi \mathrm{r}^{2} \mathrm{dr}$
$\mathrm{k}=\frac{2 \mathrm{Q}}{\pi \mathrm{R}^{4}}$
$\mathrm{QE}=\frac{1}{4 \pi \varepsilon_{0}} \frac{\mathrm{QQ}}{(2 \mathrm{a})^{2}}$
$\mathrm{R}=\mathrm{a} 8^{1 / 4}$
