(B) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{Q_{enclosed}}{\varepsilon_0}$.Here,the

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(B) According to Gauss's Law,the total electric flux $\phi$ through a closed surface is given by $\phi = \frac{Q_{enclosed}}{\varepsilon_0}$.Here,the electric field $E = A r_0$ is radially outward and uniform over the surface of a sphere of radius $r_0$.The surface area of the sphere is $S = 4\pi r_0^2$.The total electric flux $\phi$ through the sphere is $\phi = E \times S = (A r_0) \times (4\pi r_0^2) = 4\pi A r_0^3$.Equating the flux to Gauss's Law: $\frac{Q}{\varepsilon_0} = 4\pi A r_0^3$.Therefore,the enclosed charge is $Q = 4\pi \varepsilon_0 A r_0^3$.

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

Medium

The electric field in a region is radially outward with magnitude $E = A r_0$. The charge contained in a sphere of radius $r_0$ centered at the origin is

A
$4\pi \varepsilon_0 A r_0^3$
B
$4\pi \varepsilon_0 A r_0^3$
C
$\frac{4\pi \varepsilon_0 A}{r_0}$
D
$\frac{1}{4\pi \varepsilon_0} \frac{A}{r_0^3}$

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Electric Charges and Fields

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Electric Field and usage of Gauss's Law

The graphical variation of the electric field $E$ due to a uniformly charged insulating solid sphere of radius $R$ with respect to the distance $r$ from the centre $O$ is represented by:

MediumJEE MAIN 2023

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The volume charge density in a spherical ball of radius $R$ varies with distance $r$ from the centre as $\rho(r)=\rho_0\left[1-\left(\frac{r}{R}\right)^3\right]$,where $\rho_0$ is a constant. The radius at which the electric field would be maximum is

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$A$ long charged cylinder of linear charge density $\lambda$ is surrounded by a hollow coaxial conducting cylinder. What is the electric field in the space between the two cylinders?

Medium

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$A$ cube is placed inside an electric field,$\overrightarrow{E} = 150 y^2 \hat{j}$. The side of the cube is $0.5 \, m$ and it is placed in the field as shown in the figure. The charge inside the cube is $..... \times 10^{-11} \, C$.

DifficultJEE MAIN 2021

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Two infinitely long thin straight parallel wires are kept at a perpendicular distance of $2R$, having uniform linear charge densities $+\lambda$ and $-\lambda$ respectively. The magnitude of the electric field at the midpoint between the two wires will be . . . . . . .

MediumGUJCET 2026

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The electric field in a region is radially outward with magnitude $E = A r_0$. The charge contained in a sphere of radius $r_0$ centered at the origin is

Similar Questions

The graphical variation of the electric field $E$ due to a uniformly charged insulating solid sphere of radius $R$ with respect to the distance $r$ from the centre $O$ is represented by:

The volume charge density in a spherical ball of radius $R$ varies with distance $r$ from the centre as $\rho(r)=\rho_0\left[1-\left(\frac{r}{R}\right)^3\right]$,where $\rho_0$ is a constant. The radius at which the electric field would be maximum is

$A$ long charged cylinder of linear charge density $\lambda$ is surrounded by a hollow coaxial conducting cylinder. What is the electric field in the space between the two cylinders?

$A$ cube is placed inside an electric field,$\overrightarrow{E} = 150 y^2 \hat{j}$. The side of the cube is $0.5 \, m$ and it is placed in the field as shown in the figure. The charge inside the cube is $..... \times 10^{-11} \, C$.

Two infinitely long thin straight parallel wires are kept at a perpendicular distance of $2R$, having uniform linear charge densities $+\lambda$ and $-\lambda$ respectively. The magnitude of the electric field at the midpoint between the two wires will be . . . . . . .

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