The electrostatic potential inside a charged | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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

Vedclass · Accepted Answer

$ - 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

Similar Questions

The number of integral values of $k$,for which the equation $7\cos x + 5\sin x = 2k + 1$ has a solution,is

Let $z = a - \frac{i}{2}$,where $a \in R$. Then $|i + z|^2 - |i - z|^2$ is equal to

How many bridging oxygen atoms are present in $P_4O_{10}$?

Which oxide of nitrogen is obtained when ammonium nitrate is heated at $250\,^oC$?

The reduction potential of a hydrogen half-cell will be negative if:

Vedclass Test Series

Exam Paper Generator

Online Exam Module