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A point charge Q is located at the origin. A spherical Gaussian surface S₁ of radius R is centered at the origin. A second concentric spherical Gaussian surface S₂ has a radius of 3R . The region between S₁ and S₂ contains a non-uniform volume charge distribution with density = A r^2 , where A is a positive constant and r is the distance from the origin. If the electric flux through S₂ is four times the electric flux

Options

  1. A3Q 8 R
  2. B3Q 4 R
  3. CQ 8 R
  4. D4Q 3 R

Correct answer

A. 3Q 8 R

Step-by-step solution

By Gauss's law, the electric flux through a closed surface is = q_ enclosed ₀ . For the inner spherical surface S₁ of radius R , the only charge enclosed is the point charge Q at the origin. Therefore, the flux through S₁ is: ₁ = Q ₀ To find the flux through the outer surface S₂ of radius 3R , we must find the total charge enclosed by it. This includes the point charge Q and the charge distributed in the spherical shell between r = R and r = 3R . The charge q_ shell in the region between S₁ and S₂ can be found by i

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