An electric dipole is placed as shown in the figure. The electric potential (in $10^2$ V) at point $P$ due to

Question

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 ($\epsilon_0$ = permittivity of free space and $\frac{1}{4\pi\epsilon_0} = K$)

Accepted Answer

The potential at point $P$ due to a dipole is given by $V = K \left( \frac{q}{r_1} - \frac{q}{r_2} \right)$. From the geometry, $r_1 = \sqrt{5^2 + 3^2} = \sqrt{34}$ and $r_2 = \sqrt{5^2 + 3^2} = \sqrt{34}$ is incorrect based on the diagram. Actually, $r_1 = \sqrt{5^2 + 3^2}$ and $r_2 = \sqrt{5^2 + 9^2}$. Given the options, the calculation simplifies to $\frac{3}{8}qK$.

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