What is the electric flux for Gaussian surfac | vedclass

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

Question

(A) According to Gauss's law of electrostatics,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{	ext{enclosed}}}{\varep

Vedclass · Accepted Answer

$10^3 	ext{ N m}^2 	ext{ C}^{-1}$

Vedclass · Answer

(A) According to Gauss's law of electrostatics,the electric flux $\phi$ through a closed surface is given by $\phi = \frac{q_{	ext{enclosed}}}{\varepsilon_0}$.The net charge $q$ enclosed by the Gaussian surface $A$ is the sum of the individual charges:$q = q_1 + q_2 + q_3$$q = (-14 + 78.85 - 56) 	ext{ nC} = 8.85 	ext{ nC} = 8.85 	imes 10^{-9} 	ext{ C}$.The permittivity of free space is $\varepsilon_0 = 8.85 	imes 10^{-12} 	ext{ C}^2 	ext{ N}^{-1} 	ext{ m}^{-2}$.Substituting these values into the formula for electric flux:$\phi = \frac{8.85 	imes 10^{-9} 	ext{ C}}{8.85 	imes 10^{-12} 	ext{ C}^2 	ext{ N}^{-1} 	ext{ m}^{-2}}$$\phi = 10^3 	ext{ N m}^2 	ext{ C}^{-1}$.

What is the electric flux for Gaussian surface $A$ that encloses the charged particles in free space? [Given: $q_1 = -14 \text{ nC}, q_2 = 78.85 \text{ nC}, q_3 = -56 \text{ nC}$]

Similar Questions

$A$ charge $q$ is located at the centre of a cube. The electric flux through any face is

The electric flux from a cube of edge $l$ is $\phi$ for an enclosed charge. If the edge of the cube is made $\frac{2}{3} l$ and the charge enclosed in the cube is doubled,then the electric flux value will be

$A$ few electric field lines for a system of two charges $Q_1$ and $Q_2$ fixed at two different points on the $x$-axis are shown in the figure. These lines suggest that:-

Units of electric flux are

$A$ metallic spherical shell has an inner radius $R_1$ and outer radius $R_2$. $A$ charge $Q$ is placed at the centre of the spherical cavity. What will be the surface charge density on the inner surface?

Vedclass Test Series

Exam Paper Generator

Online Exam Module