Consider uniform charged shell of surface charge density s (= 3${\mathrm{\epsilon }}_0$ SI units) and a dipole of dipole moment P (= $2{\mathrm{\pi \epsilon }}_0$ SI units). Centre of the shell and the dipole lies at the origin, and dipole moment vector is along +x axis. If the field at a point on x-axis just over the shell is $\overrightarrow{{\mathrm{E}}_1}$ and that at a point on y-axis just over the shell is $\overrightarrow{{\mathrm{E}}_2}$. Find $\left|\overrightarrow{{\mathrm{E}}_1}\right|-\left|\overrightarrow{{\mathrm{E}}_2}\right|$ in N/C. Radius of shell = 50 cm.

For any point above the surface of shell the whole charge will act as a point charge at origin.

For a point on x-axis:

The field due to the shell is given by ${\mathrm{E}}_{\mathrm{s}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}.\frac{4{\mathrm{\pi R}}^2.\mathrm{\sigma }}{{\mathrm{R}}^2}$

$\Rightarrow {\mathrm{E}}_{\mathrm{S}}=3\mathrm{N}/\mathrm{C}$ towards x-axis

The field due to dipole is given by ${\mathrm{E}}_{\mathrm{D}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}.\frac{2\mathrm{P}}{{\mathrm{R}}^3}$

$\Rightarrow {\mathrm{E}}_{\mathrm{D}}=8\mathrm{N}/\mathrm{C}$ towards x-axis

So ${\mathrm{E}}_1={\mathrm{E}}_{\mathrm{S}}+{\mathrm{E}}_{\mathrm{D}}=11\mathrm{N}/\mathrm{C}$

For a point on y-axis:

Field due to the shell will be same in magnitude but different in direction.

$\Rightarrow {\mathrm{E}}_{\mathrm{S}}=3\mathrm{N}/\mathrm{C}$ towards y-axis

The field due to dipole is given by ${\mathrm{E}}_{\mathrm{D}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}.\frac{\mathrm{P}}{{\mathrm{R}}^3}$

$\Rightarrow {\mathrm{E}}_{\mathrm{D}}=4\mathrm{N}/\mathrm{C}$ towards x-axis

So, ${\mathrm{E}}_2=\sqrt{3^2+4^2}=5\mathrm{N}/\mathrm{C}$

$\left|\overrightarrow{{\mathrm{E}}_1}\right|-\left|\overrightarrow{{\mathrm{E}}_2}\right|$=6N/C