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Q.

The potential due to the electric dipole at a distance ‘ $r$ ’ from the dipole on the axial line

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a

$\frac{1}{4 π \in_{0}} \frac{4 qa}{r^{2}}$

b

$\frac{1}{4 π \in_{0}} \frac{2 qa}{r^{2}}$

c

$\frac{1}{4 π \in_{0}} \frac{qa}{r^{2}}$

d

Zero

answer is A.

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Detailed Solution

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Potential due to a dipole is given by

$\begin{aligned} V = \frac{1}{4 {πϵ}_{0}} \frac{PCos θ}{r^{2}} \\ θ = 0^{\circ} P = q \times 2 a \end{aligned}$

$\Rightarrow V = \frac{1}{4 π \in_{0}} \frac{2 qa}{r^{2}}$

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