A charge of 20 mu C is placed inside a close | vedclass

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

Question

(B) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\epsilon_0}$.Initial flux $\phi

Vedclass · Accepted Answer

$4 \ \phi$

Vedclass · Answer

(B) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\epsilon_0}$.Initial flux $\phi = \frac{20 \ \mu C}{\epsilon_0}$.After adding $80 \ \mu C$,the total charge becomes $q_{total} = 20 \ \mu C + 80 \ \mu C = 100 \ \mu C$.The new flux is $\phi' = \frac{100 \ \mu C}{\epsilon_0}$.The change in flux is $\Delta \phi = \phi' - \phi = \frac{100 \ \mu C}{\epsilon_0} - \frac{20 \ \mu C}{\epsilon_0} = \frac{80 \ \mu C}{\epsilon_0}$.Since $\phi = \frac{20 \ \mu C}{\epsilon_0}$,we can write $\Delta \phi = 4 	imes \left( \frac{20 \ \mu C}{\epsilon_0} ight) = 4 \phi$.

$A$ 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:

Similar Questions

Consider the charge configuration and spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface,the electric field will be due to:

Arrangements of charges are shown in the figure. Flux linked with the closed surfaces $P$ and $Q$ respectively are . . . . . . and . . . . . . .

$A$ point charge $q$ is placed at a distance $\frac{a}{2}$ directly above the centre of a square of side $a$. The electric flux through the square is

Gauss's law should be invalid if

Define electric flux.

Vedclass Test Series

Exam Paper Generator

Online Exam Module