The electrostatic potential inside a charged | vedclass

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Question

(A) The electric field $E$ is related to the potential $\phi$ by the relation $E = -\frac{d\phi}{dr}$.Given $\phi = ar^2 + b$,we have $E = -\frac{d}{d

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$ - 6a\varepsilon_0$

Vedclass · Answer

(A) The electric field $E$ is related to the potential $\phi$ by the relation $E = -\frac{d\phi}{dr}$.Given $\phi = ar^2 + b$,we have $E = -\frac{d}{dr}(ar^2 + b) = -2ar$.According to Gauss's Law in differential form,the charge density $ho$ is given by $ho = \varepsilon_0 (
abla \cdot E)$.In spherical coordinates,for a radial field $E_r$,the divergence is $
abla \cdot E = \frac{1}{r^2} \frac{d}{dr}(r^2 E_r)$.Substituting $E_r = -2ar$,we get $
abla \cdot E = \frac{1}{r^2} \frac{d}{dr}(r^2 (-2ar)) = \frac{1}{r^2} \frac{d}{dr}(-2ar^3) = \frac{1}{r^2} (-6ar^2) = -6a$.Therefore,the charge density is $ho = \varepsilon_0 (-6a) = -6a\varepsilon_0$.

The electrostatic potential inside a charged spherical ball is given by $\phi = ar^2 + b$,where $r$ is the distance from the centre and $a, b$ are constants. Then the charge density inside the ball is

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