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Physics Question on potential energy

Two point charges −q-q and +q+ q are located at point's (0,0,−a)(0, 0, - a) and, (0,0,a)(0, 0, a) respectively. The electric potential at a point (0,9,z)(0, 9, z), where z>az > a is

A

qa4πε0z2\frac{q_a}{4 \pi \varepsilon_0 z^2}

B

q4πε0a\frac{q}{4 \pi \varepsilon_0 a}

C

2qa4πε0z2−a2\frac{2q a}{4 \pi \varepsilon_0 z^2 - a^2}

D

2qa4πε0z2+a2\frac{2qa}{4 \pi \varepsilon_0 z^2 + a^2}

Answer

2qa4πε0z2−a2\frac{2q a}{4 \pi \varepsilon_0 z^2 - a^2}

Explanation

Solution

Potential at PP due to (+q)(+q) charge
V1=14πε0.qz−aV_{1} = \frac{1}{4 \pi \varepsilon_{0} . \frac{q}{z - a}}
Potential at PP due to (−q)(-q) charge
V2=14πε0.−qz+aV_2 = \frac{1}{4 \pi \varepsilon_0} . \frac{-q}{z + a}
Total potential at P due to (AB) electric dipole
V=V1+V2V = V_1 + V_2
=14πε0.qz−z−14πε0qz−a= \frac{1}{4 \pi \varepsilon_0} . \frac{q}{z -z} - \frac{1}{4 \pi \varepsilon_0} \frac{q}{z - a}
=q4πε0z−a−z+az−a  z+a= \frac{q}{4 \pi \varepsilon_0 } \frac{z - a - z +a}{z- a \, \, z +a}
⇒  V=2qa4πε0  z2−a2\Rightarrow \, \, V = \frac{2qa}{4 \pi \varepsilon_0 \, \, z^2 - a^2}