A square of side 20 cm is enclosed by a sphe | vedclass

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

Question

(A) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is given by $\phi = \frac{q_{enclosed}}{\epsilon_0}$,where $q_{

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(A) According to Gauss's Law,the total electric flux $\phi$ through any closed surface is given by $\phi = \frac{q_{enclosed}}{\epsilon_0}$,where $q_{enclosed}$ is the net charge enclosed by the surface.Since the square is placed inside the spherical surface and all four charges are at the corners of the square,all these charges are enclosed by the spherical surface.The net charge $q_{enclosed} = (2 - 5 - 3 + 6) 	imes 10^{-6} \ C = 0 	imes 10^{-6} \ C = 0 \ C$.Therefore,the total flux $\phi = \frac{0}{\epsilon_0} = 0 \ N \cdot m^2/C$.

Similar Questions

$A$ positive charge $Q$ is situated at the centre of a cube. The electric flux through any face of the cube is (in $SI$ units)

The figure shows two situations in which a Gaussian cube sits in an electric field. The arrows and values indicate the directions and magnitudes (in $N-m^2/C$) of the electric flux through the faces of the cube. What is the net charge (in the two situations) inside the cube?

If the number of electric lines of force emerging out of a closed surface is $1000$,then the charge enclosed by the surface is .......... $C$.

Define electric flux.

Consider four closed surfaces $S_1, S_2, S_3,$ and $S_4$ each enclosing the same charge $q_1$. Compare the electric flux through these surfaces.

Vedclass Test Series

Exam Paper Generator

Online Exam Module