Question:

A metallic sphere of radius \( R \) carrying a charge \( q \) is kept at a certain distance from another metallic sphere of radius \( R_4 \) carrying a charge \( Q \). What is the electric flux at any point inside the metallic sphere of radius \( R \) due to the sphere of radius \( R_4 \)? 

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According to Gauss’s law, the electric flux through a surface depends on the charge enclosed by that surface. Inside a conductor, there is no enclosed charge, so the electric flux is zero.
Updated On: Apr 18, 2025
  • \( \frac{Q}{4\pi \epsilon_0 R^2} \)
  • \( \frac{Q}{\epsilon_0} \)
  • \( \frac{Q}{4\pi \epsilon_0 R_4^2} \)
  • Zero
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The Correct Option is D

Solution and Explanation


To find the electric flux at any point inside the metallic sphere of radius \( R \), we need to apply Gauss's Law, which states: \[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \] where \( \Phi \) is the electric flux, \( Q_{\text{enc}} \) is the charge enclosed by the Gaussian surface, and \( \epsilon_0 \) is the permittivity of free space. For a point inside the metallic sphere of radius \( R \), the Gaussian surface inside the sphere does not enclose any charge. This is because the metallic sphere of radius \( R \) has no charge inside it; the charge \( q \) is distributed on the surface of the sphere. Thus, the electric flux at any point inside the metallic sphere of radius \( R \) due to the charge \( Q \) on the sphere of radius \( R_4 \) is zero, because the Gaussian surface inside the sphere does not enclose any charge. Therefore, the electric flux is: \[ \Phi = 0 \] Thus, the correct answer is zero.
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