(A) According to Gauss's Law,the electric flux through a closed surface is given by $\oint E \cdot dA = \frac{q_{enc}}{\varepsilon_0}$.For a Gaussian

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

(A) According to Gauss's Law,the electric flux through a closed surface is given by $\oint E \cdot dA = \frac{q_{enc}}{\varepsilon_0}$.For a Gaussian surface in the form of a sphere of radius $R$ such that $a Thus,$E(4 \pi R^2) = \frac{3Q}{\varepsilon_0}$.Solving for $E$,we get $E = \frac{3Q}{4 \pi \varepsilon_0 R^2}$.

Class 12 Physics Electric Charges and Fields Electric Field and usage of Gauss's Law

Medium

$A$ 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 center?

A
$3Q / 4 \pi \varepsilon_0 R^2$
B
$Q / 4 \pi \varepsilon_0 R^2$
C
$2Q / 4 \pi \varepsilon_0 R^2$
D
$4Q / 4 \pi \varepsilon_0 R^2$

Explore More

Browse All

Question Bank

All Subjects

Class 12 Questions

Class 12

Physics Questions

Chapter

Electric Charges and Fields

Subtopic

Electric Field and usage of Gauss's Law

Two infinite plane parallel sheets separated by a distance $d$ have equal and opposite uniform charge densities $\sigma$ and $-\sigma$. The electric field at a point between the sheets is

Easy

View Solution

As shown in the figure,a charged ball $B$ is suspended from a large charged conducting plate by a silk thread $S$ making an angle $\theta$ with the plate. The surface charge density $\sigma$ of the plate is proportional to ........

Difficult

View Solution

Let $\sigma$ be the uniform surface charge density of two infinite thin plane sheets shown in the figure. Then the electric fields in three different regions $I, II$ and $III$ are:

MediumJEE MAIN 2023

View Solution

An electrostatic field in a region is radially outward with magnitude $E = \alpha r$,where $\alpha$ is a constant and $r$ is the radial distance. The charge contained in a sphere of radius $R$ in this region (centered at the origin) is:

Medium

View Solution

$A$ solid metallic sphere has a charge $+3 Q$. 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 $B$ $(B > A)$. The electric field at a distance $R$ $(A < R < B)$ from the centre is $(\varepsilon_0 = \text{permittivity of vacuum})$

MediumMHT CET 2023

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$A$ 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 center?

Similar Questions

Two infinite plane parallel sheets separated by a distance $d$ have equal and opposite uniform charge densities $\sigma$ and $-\sigma$. The electric field at a point between the sheets is

As shown in the figure,a charged ball $B$ is suspended from a large charged conducting plate by a silk thread $S$ making an angle $\theta$ with the plate. The surface charge density $\sigma$ of the plate is proportional to ........

Let $\sigma$ be the uniform surface charge density of two infinite thin plane sheets shown in the figure. Then the electric fields in three different regions $I, II$ and $III$ are:

An electrostatic field in a region is radially outward with magnitude $E = \alpha r$,where $\alpha$ is a constant and $r$ is the radial distance. The charge contained in a sphere of radius $R$ in this region (centered at the origin) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module