Question

A cavity of radius $r/3$ is made in a sphere of radius $r$, charge $Q$. The emptied part is placed at a distance $2r$ from the centre as shown. If the charge was uniformly distributed throughout, then force between the two is :

• A. $\frac{11Q^2}{192\times 27\pi {\epsilon }_0{r}^2}$
• B. $\frac{11Q^2}{4\pi {\epsilon }_0{r}^2}$
• C. $\frac{11Q^2}{192\pi {\epsilon }_0{r}^2}$
• D. None

Answer 1

The correct option is A $\frac{11Q^2}{192\times 27\pi {\epsilon }_0{r}^2}$

Charge on the emptied part $q=\frac{Q}{r^3}(r/3{)}^3=\frac{Q}{27}$
$F=qE$ electric field due to sphere after cavity $E=[\frac{Q}{4\pi {\epsilon }_0(2r{)}^2}-\frac{Q/27}{4\pi {\epsilon }_0{\left(\frac{4r}{3}\right)}^2}]$

$E=[\frac{Q}{16\pi {\epsilon }_0{r}^2}-\frac{Q}{48\times 4\pi {\epsilon }_0{r}^2}]$
$=\frac{11Q}{16\pi {\epsilon }_0{r}^2\times 12}$

$F=qE=\frac{11Q^2}{27\times 16\pi {\epsilon }_0{r}^2\times 12}$