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NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEMedium

Question

Three charges, each +q, are placed at the corners of an isosceles triangle ABC of sides BC and AC equal to 2a. D and E are the mid points of BC and CA. The work done in taking a charge Q from D to E is:

A

\frac{3qQ}{8\pi\varepsilon_0 a}

B

\frac{qQ}{4\pi\varepsilon_0 a}

C

Zero

D

\frac{3qQ}{4\pi\varepsilon_0 a}

Step-by-Step Solution

  1. Formula for Work Done: The work done WW in moving a charge QQ from point D to point E in an electric field is given by W=Q(VE−VD)W = Q(V_E - V_D), where VEV_E and VDV_D are the electric potentials at points E and D respectively [1].
  2. Superposition Principle: The potential at any point is the algebraic sum of the potentials due to individual charges [2].
  3. Geometry Analysis:
  • The triangle ABC is isosceles with sides AC=BC=2aAC = BC = 2a.
  • Charges +q+q are placed at vertices A, B, and C.
  • D is the midpoint of BC, and E is the midpoint of AC.
  • In an isosceles triangle with equal sides AC and BC, the medians drawn to these sides are equal in length. Therefore, the distance from vertex A to the midpoint D (length AD) is equal to the distance from vertex B to the midpoint E (length BE).
  1. Potential Calculation:
  • Potential at D (VDV_D): Sum of potentials due to A, B, and C. VD=14πε0(qBD+qCD+qAD)=q4πε0(1a+1a+1AD)=q4πε0(2a+1AD)V_D = \frac{1}{4\pi\varepsilon_0} \left( \frac{q}{BD} + \frac{q}{CD} + \frac{q}{AD} \right) = \frac{q}{4\pi\varepsilon_0} \left( \frac{1}{a} + \frac{1}{a} + \frac{1}{AD} \right) = \frac{q}{4\pi\varepsilon_0} \left( \frac{2}{a} + \frac{1}{AD} \right)
  • Potential at E (VEV_E): Sum of potentials due to A, B, and C. VE=14πε0(qAE+qCE+qBE)=q4πε0(1a+1a+1BE)=q4πε0(2a+1BE)V_E = \frac{1}{4\pi\varepsilon_0} \left( \frac{q}{AE} + \frac{q}{CE} + \frac{q}{BE} \right) = \frac{q}{4\pi\varepsilon_0} \left( \frac{1}{a} + \frac{1}{a} + \frac{1}{BE} \right) = \frac{q}{4\pi\varepsilon_0} \left( \frac{2}{a} + \frac{1}{BE} \right)
  1. Conclusion: Since AD=BEAD = BE, it follows that VD=VEV_D = V_E. The potential difference ΔV=VE−VD=0\Delta V = V_E - V_D = 0. Consequently, the work done W=Q(0)=0W = Q(0) = 0.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROSTATIC POTENTIAL AND CAPACITANCE. 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.

PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEchargesplacedcornersisoscelestriangle

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