Question

The electric field at 2R from the centre of a uniformly charged non conducting sphere of radius R is E. The electric field at a distance R/2 from the centre will be
1) zero
2) 2E
3) 4E
4) 16E

Answer 1

Hello

by gauss law the field inside a hollow sphere is. zero
since the object is placed at R/2 means inside the sphere

so charge enclosed is zero hence field will be zero

option 1) is correct

Answer 2

Answer:

The electric field is 2) 2E

Given:

Radius, R = 2 R

Electric field, E = E

Solution:

According to given question, the electric field at a distance from center is given and 4 times less than distance will have electric field in relation as following:

As we know that,

$\text{Electric field}, \mathrm{E}=\frac{\mathrm{k} \mathrm{Q}}{\mathrm{R}^{2}}$

$\mathrm{E}=\frac{\frac{1}{4 \pi \epsilon_{0} \mathrm{Q}}}{R^{2}}$

$E=\frac{\frac{1}{4 \pi \epsilon_{0} Q}}{(2 R)^{2}}$

$E=\frac{Q}{16 \pi \epsilon_{0} R^{2}}$

For a non-conducting sphere with uniform charge, charge density is given by  

$\text{Charge density} =\frac{\text { charge }}{\text { volume }}$

$=\frac{\frac{Q}{4}}{\frac{4}{3 \pi R^{3}}}$

$\text{Charge density} =\frac{3 Q}{4 \pi \mathrm{R}^{3}}$

When the Radius, $\mathrm{R}^{\prime}=\frac{\mathrm{R}}{2}$

$\mathrm{Q}^{\prime}=\text { Charge density } \times \text { Volume }$

$\mathrm{Q}^{\prime}=\text { Charge density } \times \frac{4}{3} \pi\left(\frac{\mathrm{R}}{2}\right)^{3}$

$\mathrm{Q}^{\prime}=\frac{\mathrm{Q}}{\left(\frac{4}{3 \mathrm{TR}^{3}}\right) \times\left(\frac{4}{3 \pi\left(\frac{\mathrm{R}}{2}\right) 6^{3}}\right)}$

$\mathrm{Q}^{\prime}=\frac{\mathrm{Q}}{8}$

Therefore, the Electric field when radius is $\frac{\mathrm{R}}{2}$,

$\mathrm{E}^{\prime}=\frac{\frac{1}{4 \pi \epsilon_{0} \mathrm{Q}^{\prime}}}{\mathrm{R}^{\prime 2}}$

$\mathrm{E}^{\prime}=\frac{\frac{1}{4 \pi \epsilon_{0}\left(\frac{Q}{8}\right)}}{\left(\frac{R}{2}\right)^{2}}$

$\mathrm{E}^{\prime}=\frac{\mathrm{Q}}{8 \pi \epsilon_{0} \mathrm{R}^{2}}$

$\mathrm{E}^{\prime}=2\left(\frac{\mathrm{Q}}{16 \pi \epsilon_{0} \mathrm{R}^{2}}\right)$

$\mathrm{E}^{\prime}=2 \mathrm{E}$

The electric field of the sphere will be twice of electric field when the radius is $\frac{\mathrm{R}}{2}$