# (a) Draw equipotential surfaces corresponding to the electric field that uniformly increases in magnitude along with the z-directions.(b) Two charges – q and + q are located at points $(0,0,-a)$ and $(0,0,a)$. What is the electrostatic potential at the points  $(0,0,+z)$   and $(x,y,0)$?

(a)

The equipotential surfaces will be parallel to x-y plane.

(b) Electric potential at $(0,0,+z)$

Now, Electric potential at $(0,0,+z)$ due to $-q$

$V_{-}=-k\frac{q}{r}=-k\frac{q}{(z+a)}$

Where, $k=\frac{1}{4\pi \epsilon _{0}}$  and  $r=z+a$

and, Electric potential at $(0,0,+z)$ due to $+q$  charge.

$V_{+}=k\frac{q}{r}=-k\frac{q}{(z-a)}$

Where, $k=\frac{1}{4\pi \epsilon _{0}}$  and  $r=z-a$

Therefore, the net potential,

$V_{net}=V_{+}+V_{-}$

$V_{net}=\frac{kq}{z-a}+\frac{kq}{z+a}$

$=kq\left ( \frac{1}{z-a}-\frac{1}{z+a} \right )$

$=kq\left ( \frac{z+a-z+a}{z^{2}-a^{2}} \right )$

$=kq\left ( \frac{2a}{z^{2}-a^{2}} \right )$

$=\frac{2kqa}{z^{2}-a^{2}}$

At $(x,y,o)$ and as $xy$ plane is denoted by $z=0$

The distance for $-q$ and $+q$ will be the same at XY plane.

Therefore, the XY plane is an equipotential surface. Therefore the potential due to charges = 0

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