Q122 / 200 · NEET 2023 (All India) · Physics

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 $\frac{1}{4 π ε_{0}} = K$ )

A
$(\frac{8}{3}) q K$
Correct answer
B
$(\frac{3}{8}) q K$
C
$(\frac{5}{8}) q K$
D
$(\frac{8}{5}) q K$

Solution

The net electric potential at P can be calculated as follows-
$\begin{array}{rcl} V_{p} & = & \frac{K q}{(5 cm - 3 cm) \times \frac{1 m}{10^{2} cm}} - \frac{K q}{(5 cm + 3 cm) \times \frac{1 m}{10^{2} cm}} \\ = & K q \times 10^{2} \times (\frac{1}{2} - \frac{1}{8}) \\ = & \frac{3 K q}{8} \times 10^{2} V \end{array}$

Hence, the correct answer is option: 'B'.

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An electric dipole is placed as shown in the figure.The electric potential (in 102 V) at point P due to the dipole is (ε0=permittivity of free space and 14πε0=K)

Solution

Relevant Questions

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 $\frac{1}{4 π ε_{0}} = K$ )