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NEET PHYSICSGeneralMedium

Question

Two point charges −q-q and +q+q are placed at a distance of 2L2L, as shown in the figure. The magnitude of electric field intensity at a distance R(R>>L)R (R >> L) varies as:

1

1R2\frac{1}{R^2}

2

1R3\frac{1}{R^3}

3

1R4\frac{1}{R^4}

4

1R6\frac{1}{R^6}

Step-by-Step Solution

For R>>LR >> L, the arrangement acts as an electric dipole. The electric field intensity EE of a dipole at an axial or equatorial point is given by E=2p4πϵ0R3E = \frac{2p}{4\pi\epsilon_0 R^3} (for axial) or E=p4πϵ0R3E = \frac{p}{4\pi\epsilon_0 R^3} (for equatorial), where p=q(2L)p = q(2L). In both cases, E∝1R3E \propto \frac{1}{R^3}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSchargesplaceddistancefiguremagnitude

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