Write Gauss’s law and give its expression.
Write Gauss’s law and give its expression.
Similar Questions
In finding the electric field using Gauss Law the formula $|\overrightarrow{\mathrm{E}}|=\frac{q_{\mathrm{enc}}}{\varepsilon_{0}|\mathrm{A}|}$ is applicable. In the formula $\varepsilon_{0}$ is permittivity of free space, $A$ is the area of Gaussian surface and $q_{enc}$ is charge enclosed by the Gaussian surface. The equation can be used in which of the following situation?
In finding the electric field using Gauss Law the formula $|\overrightarrow{\mathrm{E}}|=\frac{q_{\mathrm{enc}}}{\varepsilon_{0}|\mathrm{A}|}$ is applicable. In the formula $\varepsilon_{0}$ is permittivity of free space, $A$ is the area of Gaussian surface and $q_{enc}$ is charge enclosed by the Gaussian surface. The equation can be used in which of the following situation?
- [JEE MAIN 2020]
A field line shown in the figure. This field line cannot represent
A field line shown in the figure. This field line cannot represent
An electric field $\overrightarrow{\mathrm{E}}=4 \mathrm{x} \hat{\mathrm{i}}-\left(\mathrm{y}^{2}+1\right) \hat{\mathrm{j}}\; \mathrm{N} / \mathrm{C}$ passes through the box shown in figure. The flux of the electric field through surfaces $A B C D$ and $BCGF$ are marked as $\phi_{I}$ and $\phi_{\mathrm{II}}$ respectively. The difference between $\left(\phi_{\mathrm{I}}-\phi_{\mathrm{II}}\right)$ is (in $\left.\mathrm{Nm}^{2} / \mathrm{C}\right)$
An electric field $\overrightarrow{\mathrm{E}}=4 \mathrm{x} \hat{\mathrm{i}}-\left(\mathrm{y}^{2}+1\right) \hat{\mathrm{j}}\; \mathrm{N} / \mathrm{C}$ passes through the box shown in figure. The flux of the electric field through surfaces $A B C D$ and $BCGF$ are marked as $\phi_{I}$ and $\phi_{\mathrm{II}}$ respectively. The difference between $\left(\phi_{\mathrm{I}}-\phi_{\mathrm{II}}\right)$ is (in $\left.\mathrm{Nm}^{2} / \mathrm{C}\right)$
- [JEE MAIN 2020]
A cubical volume is bounded by the surfaces $x =0, x = a , y =0, y = a , z =0, z = a$. The electric field in the region is given by $\overrightarrow{ E }= E _0 \times \hat{ i }$. Where $E _0=4 \times 10^4 NC ^{-1} m ^{-1}$. If $a =2 cm$, the charge contained in the cubical volume is $Q \times 10^{-14} C$. The value of $Q$ is $...........$
Take $\left.\varepsilon_0=9 \times 10^{-12} C ^2 / Nm ^2\right)$
A cubical volume is bounded by the surfaces $x =0, x = a , y =0, y = a , z =0, z = a$. The electric field in the region is given by $\overrightarrow{ E }= E _0 \times \hat{ i }$. Where $E _0=4 \times 10^4 NC ^{-1} m ^{-1}$. If $a =2 cm$, the charge contained in the cubical volume is $Q \times 10^{-14} C$. The value of $Q$ is $...........$
Take $\left.\varepsilon_0=9 \times 10^{-12} C ^2 / Nm ^2\right)$
- [JEE MAIN 2023]
Given below are two statement: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A:$ If an electric dipole of dipole moment $30 \times 10^{-5}\,Cm$ is enclosed by a closed surface, the net flux coming out of the surface will be zero.
Reason $R$ : Electric dipole consists of two equal and opposite charges.
In the light of above, statements, choose the correct answer from the options given below:
Given below are two statement: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A:$ If an electric dipole of dipole moment $30 \times 10^{-5}\,Cm$ is enclosed by a closed surface, the net flux coming out of the surface will be zero.
Reason $R$ : Electric dipole consists of two equal and opposite charges.
In the light of above, statements, choose the correct answer from the options given below:
- [JEE MAIN 2023]