Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Charges $+q$ and $-q$ are placed at points A and B, respectively, which are at a distance $2L$ apart. C is the midpoint between A and B. The work done in moving a charge $+Q$ along the semicircle CRD is:

$\frac{qQ}{4\pi\epsilon_0 L}$

$\frac{qQ}{2\pi\epsilon_0 L}$

$\frac{qQ}{6\pi\epsilon_0 L}$

$-\frac{qQ}{6\pi\epsilon_0 L}$

Step-by-Step Solution

Conservative Nature: The work done by the electrostatic force is independent of the path taken and depends only on the potential difference between the final and initial points [NCERT Class 12, Sec 2.1]. Thus, the work done moving charge $Q$ from C to D is $W = Q(V_D - V_C)$ .
Potential at C (Midpoint): Point C is equidistant ( $L$ ) from $+q$ and $-q$ . The potential $V_C$ is the sum of potentials due to individual charges: $V_C = \frac{1}{4\pi\epsilon_0} \left( \frac{+q}{L} + \frac{-q}{L} \right) = 0$
Location of D: While the diagram is not provided, the options and the nature of the problem (AIPMT 2007) imply that D is a point on the axis of the dipole, situated at distance $L$ from B (outside the dipole). Thus, the distance from $+q$ (at A) is $3L$ and from $-q$ (at B) is $L$ .
Potential at D: $V_D = \frac{1}{4\pi\epsilon_0} \left( \frac{+q}{3L} + \frac{-q}{L} \right) = \frac{q}{4\pi\epsilon_0 L} \left( \frac{1}{3} - 1 \right) = \frac{q}{4\pi\epsilon_0 L} \left( -\frac{2}{3} \right) = -\frac{q}{6\pi\epsilon_0 L}$
Work Done: $W = Q(V_D - V_C) = Q \left( -\frac{q}{6\pi\epsilon_0 L} - 0 \right) = -\frac{qQ}{6\pi\epsilon_0 L}$

Practice Mode Available

Master this Topic on Sushrut

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

Get Started