The electric field intensity on the surface of a solid charged sphere of radius $r$ and volume charge density $\rho$ is ( $\epsilon_0=$ permittivity of free space).
- Azero
- B$\frac{5 \rho r}{6 \epsilon_0}$
- C$\frac{1}{4 \pi \epsilon_0} \frac{\rho}{r}$
- D$\frac{\rho r}{3 \epsilon_0}$
Explore More
Similar Questions
Two large,thin metal plates are parallel and close to each other. Their inner faces have surface charge densities of the same sign and magnitude $17.7 \times 10^{-22} \ C/m^2$. What is the electric field $E$ in the outer region of the second plate?
Let $P(r) = \frac{Q}{\pi R^4} r$ be the charge density distribution for a solid sphere of radius $R$ and total charge $Q$. For a point $P$ inside the sphere at distance $r_1$ from the centre of the sphere,the magnitude of the electric field is
Easy
View SolutionThree infinite plane sheets carrying uniform charge densities $-\sigma, 2 \sigma, 4 \sigma$ are placed parallel to the $XZ$ plane at $Y=a, 3a, 4a$ respectively. The electric field at the point $(0, 2a, 0)$ is
Electric field at a point varies as $r^0$ for
Easy
View SolutionIf the atmospheric electric field is approximately $150 \,V/m$ and the radius of the Earth is $6400 \,km$,then the total charge on the Earth's surface is .......... coulomb.
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo