QJEEAdv07Ph4e8a: JEE Advanced 2007 Physics - Class 12 JEEAdv07Ph4e8a Solution: A spherical portion has been removed from a solid sphere having …

When cavity is not present:

Electric field inside the solid sphere

$$\overrightarrow{{\mathrm{E}}_1}=\frac{\rho }{3{\epsilon }_0}\overrightarrow{r_1}$$ ..... (i)

Solid sphere after making cavity inside solid sphere,

$$\overrightarrow{{\mathrm{E}}_2}=\frac{\rho }{3{\epsilon }_0}\overrightarrow{r_2}$$ ...... (ii)

$${\overrightarrow{E}}_{net}=\overrightarrow{E_1}-\overrightarrow{E_2}=\frac{\rho }{3{\epsilon }_0}\overrightarrow{r_1}-\frac{\rho }{3{\epsilon }_0}\overrightarrow{r_2}$$

$$=\frac{\rho }{3{\epsilon }_0}(\overrightarrow{r_1}-\overrightarrow{r_2})$$ ...... (iii)

From eq (iii), we get

$$\therefore {\mathrm{E}}_{\mathrm{net}}=\frac{\rho }{3{\epsilon }_0}(\overrightarrow{r})$$

Hence, electric field inside the emptied space (cavity) is non-zero and uniform.