(A) According to Gauss's Law,for a uniformly charged spherical shell of radius $R$ and total charge $Q$:$1$. For points outside the shell $(r > R)$,th

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

$F = \frac{1}{4 \pi \varepsilon_{0}} \frac{Qq}{r^{2}}$ for $r > R$

(A) According to Gauss's Law,for a uniformly charged spherical shell of radius $R$ and total charge $Q$:$1$. For points outside the shell $(r > R)$,the shell behaves as if all its charge is concentrated at the centre. Thus,the electric field is $E = \frac{1}{4 \pi \varepsilon_{0}} \frac{Q}{r^{2}}$. The force on a charge $q$ is $F = qE = \frac{1}{4 \pi \varepsilon_{0}} \frac{Qq}{r^{2}}$.$2$. For points inside the shell $(r Comparing these results with the given options,statement $A$ is correct.

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

Difficult

Consider the force $F$ on a charge $q$ due to a uniformly charged spherical shell of radius $R$ carrying charge $Q$ distributed uniformly over it. Which one of the following statements is true for $F$,if $q$ is placed at distance $r$ from the centre of the shell?

JEE MAIN 2020 All JEE MAIN

A
$F = \frac{1}{4 \pi \varepsilon_{0}} \frac{Qq}{r^{2}}$ for $r > R$
B
$\frac{1}{4 \pi \varepsilon_{0}} \frac{qQ}{R^{2}} > F > 0$ for $r < R$
C
$F = \frac{1}{4 \pi \varepsilon_{0}} \frac{Qq}{r^{2}}$ for all $r$
D
$F = \frac{1}{4 \pi \varepsilon_{0}} \frac{Qq}{R^{2}}$ for $r < R$

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

Previous Year

JEE MAIN 2020 Paper All JEE MAIN years →

$X$ and $Y$ are large,parallel conducting plates close to each other. Each face has an area $A$. $X$ is given a charge $Q$. $Y$ is without any charge. Points $A, B$,and $C$ are as shown in the figure.

View Solution

The electric potential $V(x)$ in a region around the origin is given by $V(x) = 4x^2 \text{ V}$. The electric charge enclosed in a cube of $1 \text{ m}$ side with its centre at the origin is (in coulomb):

MediumAIEEE 2012

View Solution

$A$ particle of charge $-q$ and mass $m$ moves in a circle of radius $r$ around an infinitely long line charge of linear density $+\lambda$. Then the time period will be given as (Consider $k$ as Coulomb's constant).

DifficultJEE MAIN 2024

View Solution

Two large,thin metal plates are parallel and close to each other. On their inner faces,the plates have surface charge densities of opposite signs and of magnitude $17.0 \times 10^{-22} \; C/m^2$. What is $E$:
$(a)$ in the outer region of the first plate,
$(b)$ in the outer region of the second plate,and
$(c)$ between the plates?

Medium

View Solution

The electrostatic potential inside a charged spherical ball is given by $\phi = ar^2 + b$,where $r$ is the distance from the centre and $a, b$ are constants. Then the charge density inside the ball is:

DifficultAIEEE 2011

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

Consider the force $F$ on a charge $q$ due to a uniformly charged spherical shell of radius $R$ carrying charge $Q$ distributed uniformly over it. Which one of the following statements is true for $F$,if $q$ is placed at distance $r$ from the centre of the shell?

Similar Questions

$X$ and $Y$ are large,parallel conducting plates close to each other. Each face has an area $A$. $X$ is given a charge $Q$. $Y$ is without any charge. Points $A, B$,and $C$ are as shown in the figure.

The electric potential $V(x)$ in a region around the origin is given by $V(x) = 4x^2 \text{ V}$. The electric charge enclosed in a cube of $1 \text{ m}$ side with its centre at the origin is (in coulomb):

$A$ particle of charge $-q$ and mass $m$ moves in a circle of radius $r$ around an infinitely long line charge of linear density $+\lambda$. Then the time period will be given as (Consider $k$ as Coulomb's constant).

The electrostatic potential inside a charged spherical ball is given by $\phi = ar^2 + b$,where $r$ is the distance from the centre and $a, b$ are constants. Then the charge density inside the ball is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module