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
Only when the Gaussian surface is an equipotential surface
- B
Only when $|\overrightarrow{\mathrm{E}}|=$ constant on the surface.
- C
For any choice of Gaussian surface.
- D
Only when the Gaussian surface is an equipotential surface and $|\overrightarrow{\mathrm{E}}|$ is constant on the surface.
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- [JEE MAIN 2015]
The electric field components in Figure are $E_{x}=\alpha x^{1 / 2}, E_{y}=E_{z}=0,$ in which $\alpha=800 \;N / C\, m ^{1 / 2} .$ Calculate
$(a)$ the flux through the cube, and
$(b)$ the charge within the cube. Assume that $a=0.1 \;m$
The electric field components in Figure are $E_{x}=\alpha x^{1 / 2}, E_{y}=E_{z}=0,$ in which $\alpha=800 \;N / C\, m ^{1 / 2} .$ Calculate
$(a)$ the flux through the cube, and
$(b)$ the charge within the cube. Assume that $a=0.1 \;m$