Flux coming out from a positive charge of 8 | 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 $\phi = \frac{q_{enclosed}}{\varepsilon_m}$,where $\varepsi

Vedclass · Accepted Answer

$\frac{2}{\varepsilon_0}$

Vedclass · Answer

(B) According to Gauss's Law,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{enclosed}}{\varepsilon_m}$,where $\varepsilon_m$ is the permittivity of the medium.Given charge $q = 8 \ C$.The permittivity of the medium is $\varepsilon_m = K \varepsilon_0$,where $K$ is the dielectric constant and $\varepsilon_0$ is the permittivity of free space.Given $K = 4$,so $\varepsilon_m = 4 \varepsilon_0$.Substituting these values into the flux formula:$\phi = \frac{8}{4 \varepsilon_0} = \frac{2}{\varepsilon_0}$.

Flux coming out from a positive charge of $8 \ C$,placed in a medium of dielectric constant $4$ is

Similar Questions

Which among the following has dimensions of charge?

The electric flux for Gaussian surface $A$ that encloses the charged particles in free space is (given $q_1 = -14\, nC$,$q_2 = 78.85\, nC$,$q_3 = -56\, nC$):

Write the $SI$ unit of electric flux.

$A$ uniformly charged conducting sphere of $2.4\; m$ diameter has a surface charge density of $80.0\; \mu C/m^2$.
$(a)$ Find the charge on the sphere.
$(b)$ What is the total electric flux leaving the surface of the sphere?

$A$ spherical rubber balloon carries a charge,uniformly distributed over its surface. As the balloon is blown up and increases in size,the total electric flux coming out of the surface

Vedclass Test Series

Exam Paper Generator

Online Exam Module