The electrostatic potential inside a charged | vedclass

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

Question

(B) The electrostatic potential $V$ inside a charged sphere is given by $V = A r^2 + B$.The electric field $E$ is related to the potential $V$ by the

Vedclass · Accepted Answer

$-6 A \varepsilon_0$

Vedclass · Answer

(B) The electrostatic potential $V$ inside a charged sphere is given by $V = A r^2 + B$.The electric field $E$ is related to the potential $V$ by the relation $E = -\frac{dV}{dr}$.Substituting the given expression for $V$:$E = -\frac{d}{dr}(A r^2 + B) = -2 A r$.According to Gauss's Law in differential form,the volume charge density $ho$ is related to the electric field $E$ by the equation $
abla \cdot E = \frac{ho}{\varepsilon_0}$.For a spherically symmetric distribution,this becomes $\frac{1}{r^2} \frac{d}{dr}(r^2 E) = \frac{ho}{\varepsilon_0}$.Substituting $E = -2 A r$:$\frac{ho}{\varepsilon_0} = \frac{1}{r^2} \frac{d}{dr}(r^2 \cdot (-2 A r)) = \frac{1}{r^2} \frac{d}{dr}(-2 A r^3) = \frac{1}{r^2} (-6 A r^2) = -6 A$.Therefore,the charge density is $ho = -6 A \varepsilon_0$.

The electrostatic potential inside a charged sphere is given as $V = A r^2 + B$,where $r$ is the distance from the centre of the sphere,$A$ and $B$ are constants. Then,the charge density in the sphere is

Similar Questions

The electric potential $V$ at any point $(x, y, z)$ (all in $m$) in space is given by $V = 4x^2 \ V$. The electric field at the point $(1 \ m, 0, 2 \ m)$ in $V/m$ is:

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:

In a certain region of space with volume $0.2 \ m^3$,the electric potential is found to be $5 \ V$ throughout. The magnitude of the electric field in this region is . . . . . . $N/C$.

Electric field in a region is given by $\vec{E} = Ax\hat{i} + By\hat{j}$, where $A = 10 \ V/m^2$ and $B = 5 \ V/m^2$. If the electric potential at a point $(10, 20)$ is $500 \ V$, then the electric potential at the origin is . . . . . . $V$.

Variation in electric potential is maximum if one goes

Vedclass Test Series

Exam Paper Generator

Online Exam Module