A metal sphere of radius R cm is charged wit | vedclass

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

Question

(B) The electric field $E$ at a distance $r$ $(r \ge R)$ from the centre of a charged sphere is given by $E = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac

Vedclass · Accepted Answer

$R \sqrt{\frac{\sigma}{\epsilon_{0} E}}$

Vedclass · Answer

(B) The electric field $E$ at a distance $r$ $(r \ge R)$ from the centre of a charged sphere is given by $E = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{q}{r^{2}}$.Surface charge density $\sigma$ is defined as $\sigma = \frac{q}{4 \pi R^{2}}$,which implies $q = 4 \pi R^{2} \sigma$.Substituting the value of $q$ into the electric field equation:$E = \frac{1}{4 \pi \epsilon_{0}} \cdot \frac{4 \pi R^{2} \sigma}{r^{2}}$$E = \frac{R^{2} \sigma}{\epsilon_{0} r^{2}}$Rearranging for $r^{2}$:$r^{2} = \frac{R^{2} \sigma}{\epsilon_{0} E}$Taking the square root on both sides:$r = R \sqrt{\frac{\sigma}{\epsilon_{0} E}}$.

$A$ metal sphere of radius $R \ cm$ is charged with $4 \pi \mu C$ and is situated in air. If $\sigma$ is the surface charge density and $E$ is the electric intensity at a distance $r$ from the centre of the sphere,then $r$ is equal to ($\epsilon_{0}$ is the permittivity of free space).

Similar Questions

Explain electric field and its source as well as magnetic field and its source.

Four point positive charges of same magnitude $Q$ are placed at four corners of a rigid square frame in $xy$ plane as shown in the figure. The plane of the frame is perpendicular to the $z$-axis. If a negative point charge is placed at a distance $z$ away from the above frame $(z << L)$,then:

Four charges,each of magnitude $q$ coulomb,are placed at points $(-1,0,0), (1,0,0), (0,-1,0)$ and $(0,1,0)$ in the $xy$-plane. The distances along the axes are measured in metres. The magnitude of the electric field at the point $(0,0,1)$ on the $Z$-axis is

If the potential at the centre of a uniformly charged ring is $V_0$,then the electric field at its centre will be (assume radius $= R$):

Vedclass Test Series

Exam Paper Generator

Online Exam Module