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JEE Advanced · Physics · 17. Electrostatics

A positive point charge of \(10^{-8} \mathrm{C}\) is kept at a distance of 20 cm from the center of a neutral conducting sphere of radius 10 cm . The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved by a distance of 10 cm further away from the center of the sphere along the radial direction. Taking \(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{Nm}^2 / \mathrm{C}^2\) (where \(\epsilon_0\) is the permittivity of free space), which of the following
statements is/are correct: :

  1. A Before the grounding, the electrostatic potential of the sphere is 450 V .
  2. B Charge flowing from the sphere to the ground because of grounding is \(5 \times 10^{-9} \mathrm{C}\).
  3. C After the grounding is removed, the charge on the sphere is \(-5 \times 10^{-9} \mathrm{C}\).
  4. D The final electrostatic potential of the sphere is 300 V .
Verified Solution

Answer & Solution

Correct Answer

(A) Before the grounding, the electrostatic potential of the sphere is 450 V .

Step-by-step Solution

Detailed explanation


Before grounding
\(\mathrm{V}_{\text {sphere }}=\left(\mathrm{V}_{\mathrm{C}}\right)_{\text {net }}=\left(\mathrm{V}_{\mathrm{C}}\right)_{\mathrm{q}}+\left(\mathrm{V}_{\mathrm{C}}\right)_{\text {ind }} \)
\( \mathrm{V}_{\text {sphere }}=\frac{\mathrm{kq}}{\ell}+0=\frac{9 \times 10^9 \times 10^{-8}}{0.2}\) \(=\frac{90}{0.2}=\frac{900}{2}=450 \mathrm{volt}\)
After grounding

\(\begin{aligned} & \frac{\mathrm{kQ}}{\ell}+\frac{\mathrm{kq}_{\mathrm{S}}}{\mathrm{R}}=\mathrm{V}_{\text {sphere }}^{\prime}=0 \\ & \mathrm{q}_{\mathrm{s}}=-\frac{\mathrm{R}}{\ell} \mathrm{q}=-\frac{1}{2} \times 10^{-8}=-5 \times 10^{-9} \\ & \mathrm{q}_{\mathrm{s}}=-5 \times 10^{-9} \text { Coulomb }\end{aligned}\)
Charge flower from sphere to ground \(=5 \times 10^{-9}\) Coulomb
After grounding is removed
\(\left(\mathrm{V}_{\text {sphere }}\right)_{\text {final }}=\frac{\mathrm{kq}}{\ell^{\prime}}+\frac{\mathrm{kq}_{\mathrm{s}}}{\mathrm{R}} \)
\( =\frac{9 \times 10^9 \times 10^2 \times 10^{-8}}{30}-\) \(\frac{9 \times 10^9 \times 5 \times 10^{-9} \times 10^2}{10} \)
\( =\frac{9 \times 1000}{30}-450=300 \mathrm{volt}-\) \(450 \mathrm{volt}=-150 \mathrm{volt}\)
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