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 _________ .
- A \(-12\)
- B \(-6\)
- C \(-9\)
- D \(-8\)
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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