91Ó°ÊÓ

Electric field is the force experienced by a unit positive test charge and is denoted by E.

Electric potential is thework doneto move unit charge against the electric field or the electric potential difference is the work done by conservative forces to move a unit positive charge and is denoted by .

Since both charges have the same sign we can conclude that electric potential at any point locates on the line connecting them always has a positive value and can't be negative or zero because the net potential is the scalar sum of first and second potential.

Both charges are positive so at point between them the electric field due to right hand charge is to left and the electric field due to left hand charge is to right so assume that the electric field is equal to zero at point x from smaller charge then using super position tells

${\mathit{E}}_{\mathbf{n}}\mathbf{=}\mathbf{0}\mathbf{⇒}\frac{\mathbf{k}\mathbf{Q}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{2}\mathbf{k}\mathbf{Q}}{{\left(d-x\right)}^{\mathbf{2}}}\mathbf{=}\mathbf{0}$ (1)

Simplify,

$\mathbf{2}{\mathit{x}}^{\mathbf{2}}\mathbf{=}{\left(d-x\right)}^{\mathbf{2}}\mathbf{⇒}\sqrt{\mathbf{2}}\mathit{x}\mathbf{=}\mathit{d}\mathbf{-}\mathit{x}\mathbf{⇒}\mathit{x}\mathbf{=}\frac{\mathbf{d}}{\mathbf{1}\mathbf{+}\sqrt{\mathbf{2}}}$

Since both charges have the opposite signs we can conclude that electric potential at point somewhere on the line connecting them is equal to zero.

So, the electric potential may equal to zero between them or beyond -Q, so try for point between them.

$\mathit{V}{\mathbf{}}_{\mathbf{n}}\mathbf{=}\mathbf{0}\mathbf{⇒}\frac{\mathbf{-}\mathbf{k}\mathbf{Q}}{\mathbf{x}}\mathbf{+}\frac{\mathbf{2}\mathbf{k}\mathbf{Q}}{\mathbf{d}\mathbf{-}\mathbf{x}}\mathbf{=}\mathbf{0}\mathbf{⇒}\mathit{x}\mathbf{=}\frac{\mathbf{d}}{\mathbf{3}}$

Try for point beyond -Q,

${\mathit{V}}_{\mathbf{n}}\mathbf{=}\mathbf{0}\mathbf{⇒}\frac{\mathbf{-}\mathbf{k}\mathbf{Q}}{\mathbf{x}}\mathbf{+}\frac{\mathbf{2}\mathbf{k}\mathbf{Q}}{\mathbf{d}\mathbf{+}\mathbf{x}}\mathbf{=}\mathbf{0}\mathbf{⇒}\mathit{x}\mathbf{=}\mathit{d}$

Similarly it's not possible for voltage to zero beyond2Q the negative charge will have a large radius unlike2Q which has a small radius so it will never be zero beyond 2Q.

Both charges have opposite sign so at point between them the electric field due to right hand charge is in the same direction with left hand charge so it's impossible to be zero, another probability is to be at x from-Q as shown in fig (2), so assume that the electric field is equal to zero at point x from smaller charge then using super position tells

role="math" localid="1664270566959" $$\begin{gathered} {\mathit{E}}_{\mathbf{n}}\mathbf{=}\mathbf{0}\mathbf{⇒}\frac{\mathbf{-}\mathbf{k}\mathbf{Q}}{{\mathbf{x}}^{\mathbf{2}}}\mathbf{+}\frac{\mathbf{2}\mathbf{k}\mathbf{Q}}{{\mathbf{(}\mathbf{d}\mathbf{+}\mathbf{x}\mathbf{)}}^{\mathbf{2}}}\mathbf{=}\mathbf{0} \\ \mathbf{⇒}\mathbf{2}{\mathit{x}}^{\mathbf{2}}\mathbf{=}{\left(d+x\right)}^{\mathbf{2}}\mathbf{⇒}\sqrt{\mathbf{2}}\mathit{x}\mathbf{=}\mathit{d}\mathbf{+}\mathit{x}\mathbf{⇒}\mathit{x}\mathbf{=}\frac{\mathbf{d}}{\sqrt{\mathbf{2}}\mathbf{-}\mathbf{1}} \end{gathered}$$

The electric field beyond 2Q can’t be zero because it’s large charge as compared with-Q and its radius.

Electric field and electric potential don’t equal to zero at same point.

Thus, $\mathit{V}\mathbf{≠}\mathbf{0}$at any point,$\mathit{E}\mathbf{=}\mathbf{0}$at $\mathit{x}\mathbf{=}\frac{\mathbf{d}}{\mathbf{1}\mathbf{+}\sqrt{\mathbf{2}}}$where x is measured from smaller charge. V=0 at $\mathit{r}\mathbf{=}\frac{\mathbf{2}\mathbf{d}}{\mathbf{3}}$measured from large charge, E=0at point $\mathit{x}\mathbf{=}\frac{\mathbf{d}}{\sqrt{\mathbf{2}}\mathbf{-}\mathbf{1}}$measured from smaller charge. Electric field and electric potential are not equal to zero at the same points.