Question

Only smart ppl can answer this so Answer this asap! ​

Answer 1

By superposition theorem the object can be replaced by,

• a sphere having same size and position of the large sphere, and having volume charge density $\small+\rho$

• a sphere having same size and position of the small sphere, and having volume charge density $\small-\rho$

as shown in the figure.

Inside the first sphere, we consider a Gaussian sphere of radius $\small\sf{r\leq 2R}$ to find electric field due to it.

Volume of Gaussian sphere is,

$\small\text{ \longrightarrow\sf{V=\dfrac{4}{3}\,\pi r^3} }$

And its surface area is,

$\small\longrightarrow\sf{A=4\pi r^2}$

The charge enclosed by Gaussian sphere is,

$\small\longrightarrow\sf{q=\rho V}$

$\small\text{ \longrightarrow\sf{q=\dfrac{4}{3}\,\pi\rho r^3} }$

By Gauss Law the electric flux density through the Gaussian surface is,

$\small\text{ \longrightarrow\displaystyle\oint\vec{\sf{E_1}}\cdot\vec{\sf{ds}}=\sf{\dfrac{q}{\epsilon_0}} }$

$\small\text{ \longrightarrow\sf{E_1\cdot4\pi r^2=\dfrac{4\pi\rho r^3}{3\epsilon_0}} }$

$\small\text{ \longrightarrow\sf{E_1=\dfrac{\rho r}{3\epsilon_0}} }$

Now the potential difference between A and B due to this electric field is,

$\small\text{ \longrightarrow\sf{\displaystyle V_{A1}-V_{B1}=-\int\limits_{2R}^RE\,dr} }$

$\small\text{ \longrightarrow\sf{\displaystyle V_{A1}-V_{B1}=-\dfrac{\rho}{3\epsilon_0}\int\limits_{2R}^Rr\,dr} }$

$\small\text{ \longrightarrow\sf{V_{A1}-V_{B1}=-\dfrac{\rho}{6\epsilon_0}\left[r^2\right]_{2R}^R} }$

$\small\text{ \longrightarrow\sf{V_{A1}-V_{B1}=\dfrac{\rho R^2}{2\epsilon_0}} }$

In the second sphere, A and B are diametrically opposite points wrt the center O' so the potential difference between them is zero.

$\small\text{ \longrightarrow\sf{V_{A2}-V_{B2}=0} }$

Now the net potential difference is,

$\small\text{ \longrightarrow\sf{V_A-V_B=\dfrac{\rho R^2}{2\epsilon_0}+0} }$

$\small\text{ \longrightarrow\sf{V_A-V_B=\dfrac{1}{2}\cdot\dfrac{\rho}{\epsilon_0}\cdot R^2} }$

Given that,

• $\small\text{ \sf{\dfrac{\rho}{\epsilon_0}=16} }$

• $\small\text{ \sf{R=\dfrac{3}{2}} }$

Hence,

$\small\text{ \longrightarrow\sf{V_A-V_B=\dfrac{1}{2}\cdot16\cdot\left(\dfrac{3}{2}\right)^2} }$

$\small\text{ \longrightarrow\sf{\underline{\underline{V_A-V_B=18\,V}}} }$