(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)?

 

 

 

 
 
 
 
 

Answers (1)

(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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