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Electrostatics question

2023 · Q163
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Electrostatics question

2023 · Q163

NEETPhysicsElectrostaticsMCQ+4 / −1

If a conducting sphere of radius R\mathrm{R}R is charged. Then the electric field at a distance r(r>R)\mathrm{r}(\mathrm{r} > \mathrm{R})r(r>R) from the centre of the sphere would be, (V=(\mathrm{V}=(V= potential on the surface of the sphere)

  1. A
    rVR2\frac{r V}{R^2}R2rV​
  2. B
    R2Vr3\frac{R^2 V}{r^3}r3R2V​
  3. C
    RVr2\frac{R V}{r^2}r2RV​
  4. D
    Vr\frac{\mathrm{V}}{\mathrm{r}}rV​
View written solutionFree

Correct answer: C

$$ \begin{aligned} & \therefore V=\frac{K Q}{R} \\ & E=\frac{K Q}{r^2} \\ & E=\frac{V R}{r^2} \end{aligned} $$

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