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 $\dfrac{1}{4\pi{ε}_{0}} = K$ )

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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 1 4π ε _ 0 = K )

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Correct answer: B. ((3)/(8))qK

The net electric potential at P can be calculated as follows-
$\begin{cases}{V}_{p} & = & \dfrac{Kq}{\left(5cm - 3cm\right) \times \dfrac{1m}{{10}^{2}cm}} - \dfrac{Kq}{\left(5cm + 3cm\right) \times \dfrac{1m}{{10}^{2}cm}} \\ & = & Kq \times {10}^{2} \times \left(\dfrac{1}{2} - \dfrac{1}{8}\right) \\ & = & \dfrac{3Kq}{8} \times {10}^{2}V\end{cases}$

An electric dipole is placed as shown in the figure. The electric...

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