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JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance

The electrostatic potential in a charged spherical region of radius \(r\) varies as \(V=a r^3+b\), where \(a\) and b are constants. The total charge in the sphere of unit radius is \(\alpha \times \pi a \in_0\). The value of \(\alpha\) is _________ .

  1. A \(-12\)
  2. B \(-6\)
  3. C \(-9\)
  4. D \(-8\)
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Answer & Solution

Correct Answer

(A) \(-12\)

Step-by-step Solution

Detailed explanation

\( v=ar^{3}+b \) \( E=-\frac{dv}{dr}=-3ar^{2} \) \( \phi_{closed}=\frac{q_{cnc}}{\epsilon_{0}} \) \( q_{cnc}=\epsilon_{0}.E.A \) \( =\epsilon_{0}(-3a.(1)^{2})4\pi(1)^{2} \) \( =-12\pi a\epsilon_{0} \) \(\therefore x =-12\)
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