Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

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 $\frac{1}{4\pi\epsilon_0} = K$ )

$\frac{3}{8}qK$

$\frac{5}{8}qK$

$\frac{8}{5}qK$

$\frac{8}{3}qK$

Step-by-Step Solution

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 $\frac{3}{8}qK$ .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started