The electrostatic potential inside a charged | vedclass

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

Question

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

Vedclass · Accepted Answer

$-6 a\varepsilon_0$

Vedclass · Answer

(D) The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dr}$.Given $V = b - ar^2$,we have $E = -\frac{d}{dr}(b - ar^2) = 2ar$.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)$.Substituting $E = 2ar$,we get $
abla \cdot E = \frac{1}{r^2} \frac{d}{dr}(r^2 \cdot 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 $V = b - ar^2$,where $r$ is the distance from the centre; $a$ and $b$ are constants. Then,the charge density inside the ball is:

Similar Questions

The electric potential at a point in free space due to charge $Q$ coulomb is $V = Q \times 10^{11} \, V$. The electric field at that point is:

$A$ uniform electric field of $400 \, V/m$ is directed $45^\circ$ above the $x$-axis. The potential difference $V_A - V_B$ is: (Coordinates are in cm) (in $, V$)

Due to some charges on the $x$-axis,the potential at a point on the $x$-axis is given by $V(x) = 20/(x^2 - 4) \ V$. The electric field at $x = 4 \ \mu m$ is given by:

Assume that an electric field $\overrightarrow{E} = 30x^2 \hat{i}$ exists in space. Then the potential difference $V_A - V_O$,where $V_O$ is the potential at the origin and $V_A$ is the potential at $x = 2 \, m$,is: (in $, V$)

$A$ small particle carrying a negative charge of $1.6 \times 10^{-19} \text{ C}$ is suspended in equilibrium between two horizontal metal plates $8 \text{ cm}$ apart having a potential difference of $980 \text{ V}$ across them. Find the mass of the particle. $[g = 9.8 \text{ m/s}^2]$

Vedclass Test Series

Exam Paper Generator

Online Exam Module