A point charge +10 ; mu C is at a distance 5 | vedclass

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

Question

(C) The square can be considered as one face of a cube of edge $10 \; cm$ with the charge $q$ placed at its centre. According to Gauss's theorem,the t

Vedclass · Accepted Answer

$1.88 	imes 10^{5} \; N \; m^{2} \; C^{-1}$

Vedclass · Answer

(C) The square can be considered as one face of a cube of edge $10 \; cm$ with the charge $q$ placed at its centre. According to Gauss's theorem,the total electric flux through a closed surface is $\phi_{	ext{Total}} = \frac{q}{\varepsilon_{0}}$.Since the charge is at the centre of the cube,the total flux is distributed equally among its six faces. Therefore,the electric flux through one face (the square) is:$\phi = \frac{\phi_{	ext{Total}}}{6} = \frac{1}{6} \cdot \frac{q}{\varepsilon_{0}}$Given:$q = 10 \; \mu C = 10 	imes 10^{-6} \; C$$\varepsilon_{0} = 8.854 	imes 10^{-12} \; C^{2} \; N^{-1} \; m^{-2}$Substituting the values:$\phi = \frac{1}{6} 	imes \frac{10 	imes 10^{-6}}{8.854 	imes 10^{-12}}$$\phi = \frac{1}{6} 	imes 1.129 	imes 10^{6}$$\phi \approx 1.88 	imes 10^{5} \; N \; m^{2} \; C^{-1}$

$A$ point charge $+10 \; \mu C$ is at a distance $5 \; cm$ directly above the centre of a square of side $10 \; cm$,as shown in the figure. What is the magnitude of the electric flux through the square?

Similar Questions

An infinite line charge is at the axis of a cylinder of length $1 \,m$ and radius $7 \,cm$. If the electric field at any point on the curved surface of the cylinder is $250 \,NC^{-1}$,then the net electric flux through the cylinder is ............ $Nm^2C^{-1}$.

In a region,the electric field is given by $\overrightarrow{E}=(3 \hat{i}+5 \hat{j}+7 \hat{k}) \text{ NC}^{-1}$. The electric flux through a surface of area $3 \text{ m}^2$ in the $yz$-plane is (in $SI$ units):

An electric field,$\overrightarrow{E} = \frac{2 \hat{i} + 6 \hat{j} + 8 \hat{k}}{\sqrt{6}} \ V/m$,passes through a surface of $4 \ m^2$ area having a unit normal vector $\hat{n} = \left( \frac{2 \hat{i} + \hat{j} + \hat{k}}{\sqrt{6}} \right)$. The electric flux through that surface is:

Five point charges $\frac{1}{\pi}, \frac{2}{\pi}, \frac{3}{\pi}, \frac{4}{\pi}$ and $\frac{-5}{\pi} \ nC$ are located inside a pyramid. The total electric flux through the surface of the pyramid is

When is electric flux said to be positive,negative,or zero?

Vedclass Test Series

Exam Paper Generator

Online Exam Module