$A$ solid metallic sphere has a charge $+3Q$. Concentric with this sphere is a conducting spherical shell having charge $-Q$. The radius of the sphere is $a$ and that of the spherical shell is $b$ $(b > a)$. What is the electric field at a distance $R$ $(a < R < b)$ from the centre?
- A$\frac{Q}{2\pi \varepsilon_0 R}$
- B$\frac{3Q}{2\pi \varepsilon_0 R}$
- C$\frac{3Q}{4\pi \varepsilon_0 R^2}$
- D$\frac{4Q}{4\pi \varepsilon_0 R^2}$
Explore More
Similar Questions
As shown in the figure,a cuboid lies in a region with an electric field $\vec{E} = 2x^2 \hat{i} - 4y \hat{j} + 6 \hat{k} \; N/C$. The magnitude of the charge within the cuboid is $n \varepsilon_0 \; C$. The value of $n$ is $............$ (if the dimensions of the cuboid are $1 \times 2 \times 3 \; m^3$)
Electric field inside a uniformly charged sphere of radius $R$ is (where $r$ is the distance from the centre,$r < R$):
Medium
View SolutionThe electric field due to an infinitely long thin straight wire with uniform linear charge density of $2.5 \times 10^{-7} \ Cm^{-1}$ at a radial distance of $x$ from the wire is $7.5 \times 10^4 \ NC^{-1}$. Then $x=$ (in $cm$)
Two concentric conducting thin spherical shells $A$ and $B$ having radii $r_A$ and $r_B$ $(r_B > r_A)$ are charged to $Q_A$ and $-Q_B$ $(|Q_B| > |Q_A|)$. The electric field along a line passing through the centre is:
DifficultAIIMS 2005
View SolutionGauss's law can help in the easy calculation of the electric field due to:
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