Explanation:
(a) Let us assume that the potential at point x is equal to zero.
Then,
[tex]k \frac{2q}{x} - \frac{k(-d)}{(d - x)}[/tex] = 0
[tex]\frac{2}{x} + \frac{1}{(d- x)}[/tex] = 0
[tex]\frac{2}{x} = \frac{-1}{(d-x)}[/tex]
2d - 2x = -x
-x + 2x = 2d
or, x = 2d
Therefore, at a point x = 2d the electric potential is equal to zero.
(b) There will be only one point where an electric potential is zero and that point is x = 2d.
(c) Let us assume that [tex]E_{A}[/tex] and [tex]E_{B}[/tex] are electric fluids because of the charge +2q and -q respectively.
So, there is only one point where the charge can be zero and that point is D.
