Question

The electrostatic potential due to certain charge distribution is given by V (x, y, z) = - $\frac{V_o}{a^4}$ (x + y + z) xyz. In Column I, the values of electric field (in terms of $\frac{V_o}{a}$) and charge density (in terms of $\frac{V_o{\epsilon }_o}{a^2}$ are given. In Column II, different points in space are given. Each value in Column I may correspond to one or more quantities in Column II.

Answer 1

We know that,
$\overrightarrow{E}=-▽\varphi$
$\overrightarrow{▽}.\overrightarrow{E}=\rho /{ϵ}_0$
Therefore,
$\varphi =-V_0/a^4\times xyz(x+y+z)$
$\Rightarrow \overrightarrow{E}=V_0/a^4\times \left(\right(2xyz+y^{2}z+z^{2}y)\hat{i}+(2xyz+x^{2}z+z^{2}x)\hat{j}+(2xyz+x^{2}y+x{y}^2\left)\hat{k}\right)$
$\Rightarrow \rho ={ϵ}_0\times V_0/a^4\times (2yz+2xz+2xy)$
Now,
$\rho (a,a,a)={ϵ}_0\times V_0/a^4\times 6a^2=6{ϵ}_0{V}_0/a^2$
$E(0,0,a)=V_0/a^4\times \left(0\right)=0$
$\rho (0,a,a)={ϵ}_0\times V_0/a^4\times 2a^2=2{ϵ}_0{V}_0/a^2$
$E(a,a,a)=V_0/a^4\times 4a^3\hat{i}+4a^3\hat{j}+4a^3\hat{k}$
$E(0,a,a)=V_0/a^4\times 2a^3\hat{i}$