An electric dipole is placed as shown in the figure. The electric potential (in V) at point due to the dipole is [2023]

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Solution

Q.
An electric dipole is placed as shown in the figure. The electric potential (in $10^{2}$ V) at point $P$ due to the dipole is $(ε_{0} = permittivity of free space and \frac{1}{4 π ε_{0}} = K)$ [2023]

1 $(\frac{8}{5}) q K$

2 $(\frac{8}{3}) q K$

3 $(\frac{3}{8}) q K$

4 $(\frac{5}{8}) q K$

Ans.
(3)

Here, the dipole distance is nearly equal to the distance between the point $P$ thus, we cannot apply the dipole formula directly. We have to calculate the potential. Hence, electric potential due to charge $q$ is given by,

due to individual charges

Potential due to charge $' q'$ at point $' P'$ is

$V_{q} = \frac{1}{4 π ε_{0}} \frac{q}{r} = K q \frac{1}{(R) \times 10^{- 2}} = \frac{K q (10^{2})}{2}$ ...(i)

Potential due to charge $' - q'$ at point $' P'$ is,

   $V_{- q} = \frac{1}{4 π ε_{0}} \frac{(- q)}{(8 \times 10^{- 2})} = \frac{- K q (10^{2})}{8}$ ...(ii)

From equations (i) and (ii), we get

   $V_{net} = K q (10^{2}) (\frac{1}{2} - \frac{1}{8})$

   $V_{n e t} = (\frac{3}{8}) q k (10^{2}) V$

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