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(D) According to Gauss's law,the net electric flux $\phi_{net}$ through a closed surface is equal to the total charge $q$ enclosed by the surface divi

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$\varepsilon_0(\phi_2-\phi_1)$

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(D) According to Gauss's law,the net electric flux $\phi_{net}$ through a closed surface is equal to the total charge $q$ enclosed by the surface divided by the permittivity of free space $\varepsilon_0$.$\phi_{net} = \frac{q}{\varepsilon_0}$Here,the flux entering the surface is $\phi_1$ (which is negative) and the flux leaving the surface is $\phi_2$ (which is positive).Therefore,the net flux is $\phi_{net} = \phi_2 - \phi_1$.Substituting this into Gauss's law:$\phi_2 - \phi_1 = \frac{q}{\varepsilon_0}$$q = \varepsilon_0(\phi_2 - \phi_1)$.

If the electric flux entering and leaving an enclosed surface are $\phi_1$ and $\phi_2$ respectively,the electric charge inside the surface will be

Similar Questions

Assertion: Four point charges $q_1, q_2, q_3$ and $q_4$ are as shown in the figure. The flux over the shown Gaussian surface depends only on charges $q_1$ and $q_2$.
Reason: Electric field at all points on the Gaussian surface depends only on charges $q_1$ and $q_2$.

The figure shows electric field lines due to a charge configuration. From this,we conclude that:

$A$ circular plate sheet of radius $10 \,cm$ is placed in a uniform electric field of $2 \sqrt{3} \times 10^5 \,NC^{-1}$, making an angle of $60^{\circ}$ with the field. Find the electric flux through the sheet.

In a region,the intensity of an electric field is given by $E = 2i + 3j + k$ in $NC^{-1}$. The electric flux through a surface $S = 10i \ m^2$ in the region is

For a given surface,the Gauss's law is stated as $\oint {E \cdot ds} = 0$. From this,we can conclude that:

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