The electrostatic potential in a charged sphe | vedclass

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(A) The electrostatic potential is given by $V = ar^3 + b$. The electric field $E$ is related to the potential by $E = -\frac{dV}{dr}$. Differentiatin

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$-12$

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(A) The electrostatic potential is given by $V = ar^3 + b$. The electric field $E$ is related to the potential by $E = -\frac{dV}{dr}$. Differentiating $V$ with respect to $r$, we get $E = -\frac{d}{dr}(ar^3 + b) = -3ar^2$. According to Gauss's Law, the total charge $q_{enc}$ enclosed in a sphere of radius $r=1$ is given by $q_{enc} = \epsilon_0 \oint E \cdot dA$. For a spherical surface of radius $r=1$, the area $A = 4\pi r^2 = 4\pi(1)^2 = 4\pi$. Substituting the values at $r=1$, $E = -3a(1)^2 = -3a$. Thus, $q_{enc} = \epsilon_0 	imes (-3a) 	imes 4\pi = -12\pi a \epsilon_0$. Comparing this with the given expression $\alpha 	imes \pi a \epsilon_0$, we find $\alpha = -12$.

The electrostatic potential in a charged spherical region of radius $r$ varies as $V = ar^3 + b$, where $a$ and $b$ are constants. The total charge in the sphere of unit radius is $\alpha \times \pi a \epsilon_0$. The value of $\alpha$ is . . . . . . .

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