A charge Q ; mu C is placed at the centre of | vedclass

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

Question

(D) According to Gauss's Law,the total electric flux $\varphi_{net}$ through a closed surface is equal to the total charge enclosed divided by the per

Vedclass · Accepted Answer

$\frac{Q}{6 \varepsilon_{0}} 	imes 10^{-6}$

Vedclass · Answer

(D) According to Gauss's Law,the total electric flux $\varphi_{net}$ through a closed surface is equal to the total charge enclosed divided by the permittivity of free space $\varepsilon_{0}$.For a cube with a charge $Q \; \mu C$ placed at its centre,the total flux is $\varphi_{net} = \frac{Q 	imes 10^{-6}}{\varepsilon_{0}}$.Since a cube has $6$ identical faces and the charge is at the centre,the flux is distributed equally among all faces.Therefore,the flux through any one face is $\varphi_{face} = \frac{\varphi_{net}}{6} = \frac{Q 	imes 10^{-6}}{6 \varepsilon_{0}} = \frac{Q}{6 \varepsilon_{0}} 	imes 10^{-6}$.

$A$ charge $Q \; \mu C$ is placed at the centre of a cube. The flux coming out from any one of its faces will be:

Similar Questions

$A$ charge of $20 \ \mu C$ is placed inside a closed surface. If an additional charge of $80 \ \mu C$ is placed inside the surface,the change in flux is:

The given figure shows the electric field lines due to two charges $q_1$ and $q_2$. What are the signs of the two charges?

Discuss some points about Gauss's law.

An arbitrary surface encloses a dipole. What is the electric flux through this surface?

Find the electric flux through the shaded face $BCGF$ of the cube,as shown in the figure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module