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NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEMedium

Question

Four equal charges Q are placed at the four corners of a square of each side is ‘a’. Work done in removing a charge –Q from its centre to infinity is

A

0

B

\frac{\sqrt{2}Q^2}{4\pi\epsilon_0 a}

C

\frac{\sqrt{2}Q^2}{\pi\epsilon_0 a}

D

\frac{Q^2}{2\pi\epsilon_0 a}

Step-by-Step Solution

  1. Identify Geometry: The distance rr from any corner of the square (side aa) to the center is half the diagonal. r=a22=a2r = \frac{a\sqrt{2}}{2} = \frac{a}{\sqrt{2}}.
  2. Calculate Potential at Center (VcV_c): The electric potential is a scalar quantity, so we sum the potentials due to the four charges +Q+Q at the corners [Source 157]. Vc=4×14πϵ0Qr=1πϵ0Q(a/2)=2Qπϵ0aV_c = 4 \times \frac{1}{4\pi\epsilon_0} \frac{Q}{r} = \frac{1}{\pi\epsilon_0} \frac{Q}{(a/\sqrt{2})} = \frac{\sqrt{2}Q}{\pi\epsilon_0 a}
  3. Calculate Potential Energy (UU): The potential energy of the charge −Q-Q at the center is U=(−Q)VcU = (-Q)V_c [Source 36, 161]. Ucenter=−Q(2Qπϵ0a)=−2Q2πϵ0aU_{center} = -Q \left( \frac{\sqrt{2}Q}{\pi\epsilon_0 a} \right) = -\frac{\sqrt{2}Q^2}{\pi\epsilon_0 a}
  4. Calculate Work Done (WW): The work done to remove the charge to infinity (where potential energy is zero) is the change in potential energy. W=Ufinal−Uinitial=0−Ucenter=2Q2πϵ0aW = U_{final} - U_{initial} = 0 - U_{center} = \frac{\sqrt{2}Q^2}{\pi\epsilon_0 a}

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 CAPACITANCEchargesplacedcornerssquareremoving

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