Question

Figure shows, in cross section, two solid spheres with uniformly distributed charge throughout their volumes. Each has radius R. Points P lies on a line connecting the centres of the spheres, at radial distance $\frac{R}{2.00}$ from the centre of sphere 1. If the net electric field at point P is zero, what is the ratio $\frac{q_2}{q_1}$ of the total charges?

Answer 1

Step 1: Electric field due to sphere 1
We know that electric field at a point P lying inside the sphere 1 at a distance $r_1$ enclosing charge $q_1$ is
$E_1=\frac{q_1{r}_1}{4\pi {\epsilon }_0{R}^3}$
Here,
$r_1=\frac{R}{2}$
$E_1=\frac{|q_1|}{4\pi {\epsilon }_0{R}^3}\left(\frac{R}{2}\right)$
$=\frac{1}{2}\frac{|q_1|}{4\pi {\epsilon }_0{R}^2}$ ....(1)

Step 2: Electric field due to sphere 2
We know that electric field at a point P lying outside the sphere 2 at a distance r enclosing charge $q_2$ is

$E_2=\frac{q_2}{4\pi {\epsilon }_0{r}^2}$
Here,
$r=\frac{3R}{2}$
$E_2=\frac{|q_2|}{4\pi {\epsilon }_0{r}^2}$
$=\frac{|q_2|}{4\pi {\epsilon }_0(1.50R{)}^2}$ ....(2)

Step 3: Take ratio
From above equations (1) and (2) , we have
$\frac{E_2}{E_1}=\frac{q_2}{q_1}=\frac{9}{8}$
Hence, option (a) is correct.