Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

An electric dipole is placed as shown in the figure. The electric potential (in 10² V) at the point P due to the dipole is: (ε₀ = permittivity of free space and 1/4\pi ε₀ = k)

(8/3) qk

(3/8) qk

(5/8) qk

(8/5) qk

Step-by-Step Solution

Principle of Superposition: The electric potential ( $V$ ) at any point due to a system of charges is the algebraic sum of the potentials due to the individual charges. For a dipole consisting of charges $+q$ and $-q$ , the potential at point P is given by: $V_P = V_{+q} + V_{-q} = k \frac{q}{r_1} + k \frac{-q}{r_2} = kq \left( \frac{1}{r_1} - \frac{1}{r_2} \right)$ , where $r_1$ and $r_2$ are the distances of point P from $+q$ and $-q$ respectively.
Analysis of Probable Answer: The correct option is given as $\frac{3}{8}qk$ . This implies that the geometric factor $\left( \frac{1}{r_1} - \frac{1}{r_2} \right)$ must equal $\frac{3}{8}$ .
Hypothetical Geometry: For example, if the distance from $+q$ to P is 2 units ( $r_1 = 2$ ) and from $-q$ to P is 8 units ( $r_2 = 8$ ), then $\frac{1}{2} - \frac{1}{8} = \frac{4}{8} - \frac{1}{8} = \frac{3}{8}$ . Without the figure, the exact distances cannot be verified, but the method remains the scalar addition of potentials.

Practice Mode Available

Master this Topic on Sushrut

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

Get Started