The volume charge density of a sphere of radi | vedclass

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

Question

(C) The number of electric field lines per unit area is equivalent to the electric field intensity $E$ at that surface.For a uniformly charged sphere

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(C) The number of electric field lines per unit area is equivalent to the electric field intensity $E$ at that surface.For a uniformly charged sphere of radius $R$ and volume charge density $ho$,the electric field at the surface $(r = R)$ is given by Gauss's Law as $E = \frac{q}{4 \pi \epsilon_{0} R^{2}}$.Since $q = ho 	imes 	ext{Volume} = ho 	imes \frac{4}{3} \pi R^{3}$,we substitute this into the expression for $E$:$E = \frac{ho 	imes \frac{4}{3} \pi R^{3}}{4 \pi \epsilon_{0} R^{2}} = \frac{ho R}{3 \epsilon_{0}}$.Given $ho = 2 \, \mu C \, m^{-3} = 2 	imes 10^{-6} \, C \, m^{-3}$,$R = 6 \, m$,and $\epsilon_{0} = 8.85 	imes 10^{-12} \, C^{2} N^{-1} m^{-2}$.$E = \frac{2 	imes 10^{-6} 	imes 6}{3 	imes 8.85 	imes 10^{-12}} = \frac{12 	imes 10^{-6}}{26.55 	imes 10^{-12}} \approx 0.4519 	imes 10^{6} 	imes 10^{6} = 0.4519 	imes 10^{12} \, N C^{-1}$.Rounding to the nearest integer as per the options,we get $45 	imes 10^{10} \, N C^{-1}$.

Similar Questions

In a uniformly charged sphere of total charge $Q$ and radius $R$,the electric field $E$ is plotted as a function of distance $r$ from the center. The graph which would correspond to the above is:

Mention applications of Gauss's law.

The electric field $\vec{E} = E_0 y \hat{j}$ acts in a space where a cylinder of radius $r$ and length $l$ is placed with its axis parallel to the $y$-axis. The charge inside the volume of the cylinder is:

Find the electric field due to an infinitely long charged cylinder of radius $R$ having a linear charge density $\lambda$ at a distance of $R/2$ from its axis.

Two long thin charged rods with charge density $\lambda$ each are placed parallel to each other at a distance $d$ apart. The force per unit length exerted on one rod by the other will be (where $k = \frac{1}{4 \pi \varepsilon_0}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module