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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}
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