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NEET PHYSICSELECTROSTATIC POTENTIAL AND CAPACITANCEMedium

Question

The electric potential at a point in free space due to a charge QQ coulomb is Q×1011Q \times 10^{11} V. The electric field at that point is:

A

4πϵ0Q×1022 V/m4\pi\epsilon_0 Q \times 10^{22} \text{ V/m}

B

12πϵ0Q×1020 V/m12\pi\epsilon_0 Q \times 10^{20} \text{ V/m}

C

4πϵ0Q×1020 V/m4\pi\epsilon_0 Q \times 10^{20} \text{ V/m}

D

12πϵ0Q×1022 V/m12\pi\epsilon_0 Q \times 10^{22} \text{ V/m}

Step-by-Step Solution

  1. Formulas: Electric potential VV due to a point charge QQ at distance rr: V=14πϵ0QrV = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} [NCERT Class 12, Sec 2.3, Eq 2.8]. Electric field magnitude EE due to a point charge QQ at distance rr: E=14πϵ0Qr2E = \frac{1}{4\pi\epsilon_0} \frac{Q}{r^2} [NCERT Class 12, Sec 1.8, Eq 1.6].
  • Relationship: E=VrE = \frac{V}{r}.
  1. Calculation: Given V=Q×1011V = Q \times 10^{11} Volts. Substitute into the potential formula to find rr: Q×1011=14πϵ0QrQ \times 10^{11} = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} r=14πϵ0×1011r = \frac{1}{4\pi\epsilon_0 \times 10^{11}}
  • Substitute rr into the field relationship E=V/rE = V/r: E=Q×101114πϵ0×1011E = \frac{Q \times 10^{11}}{\frac{1}{4\pi\epsilon_0 \times 10^{11}}} E=(Q×1011)×(4πϵ0×1011)E = (Q \times 10^{11}) \times (4\pi\epsilon_0 \times 10^{11}) E=4πϵ0Q×1022 V/mE = 4\pi\epsilon_0 Q \times 10^{22} \text{ V/m}

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 CAPACITANCEelectricpotentialchargecoulombelectric

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