Along the X-axis, three charges $\frac{q}{2}, - q$ and $\frac{q}{2}$ are placed at x = 0, x = a and x = 2a respectively. The resultant electric potential at x = a+r (if, a << r) is :

( $ϵ_{0}$ is the permittivity of free space)

Q. Two equal negative point charges

- q

are fixed at points

(0, a)

and

(0, - a)

on the

y-

axis. A positive point charge

Q

is in rest at point

(2 a, 0)

on the

x-

axis. The net electric field on

Q

due to both

- q

charges will be

(\frac{1}{4 π ϵ_{0}} = k)

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Two point charge -q and +q/2 are situated at the origin and the point (a,0,0) respectively. The point along the X-axis where the electric field vanishes is

The correct option is A x=a√2Two point charges −q and +q/2 point (a,0,0) respectively.The electric field x=a√12+12=a√2

The correct option is A $x = \frac{a}{\sqrt{2}}$
Two point charges $- q$ and $+ q / 2$ point $(a, 0, 0)$ respectively.
The electric field
$x = \frac{a}{\sqrt{1^{2} + 1^{2}}} = \frac{a}{\sqrt{2}}$