A point charge $+10\; \mu \,C$ is a distance $5 cm$ directly above the centre of a square of side $10 \;cm ,$ as shown in Figure. What is the magnitude of the electric flux through the square?
The square can be considered as one face of a cube of edge $10 \,cm$ with a centre where charge $q$ is placed. According to Gauss's theorem for a cube, total electric flux is through all its six faces.
$\phi_{\text {Total}}=\frac{q}{\varepsilon_{0}}$
Hence, electric flux through one face of the cube i.e., through the square is
$\phi=\frac{\phi_{\text {Total}}}{6}=\frac{1}{6} \cdot \frac{q}{\varepsilon_{0}}$
Where, $\varepsilon_{0}=$ Permittivity of free space $=8.854 \times 10^{-12}\, N ^{-1} \,C ^{2} \,m ^{-2}$
$q=10\, \mu \,C=10 \times 10^{-6} \,C$
$=1.88 \times 10^{5} \,N \,m ^{2} \,C ^{-1}$
$\therefore \phi=\frac{1}{6} \cdot \frac{10 \times 10^{-6}}{8.854 \times 10^{-12}}$
Therefore, electric flux through the square is $1.88 \times 10^{5} \;N \;m ^{2} \,C ^{-1}$
Given below are two statement: one is labelled as Assertion $A$ and the other is labelled as Reason $R$.
Assertion $A:$ If an electric dipole of dipole moment $30 \times 10^{-5}\,Cm$ is enclosed by a closed surface, the net flux coming out of the surface will be zero.
Reason $R$ : Electric dipole consists of two equal and opposite charges.
In the light of above, statements, choose the correct answer from the options given below:
Draw electric field by positive charge.
Obtain Gauss’s law from the flux associated with a sphere of radius $\mathrm{'r'}$ and charge $\mathrm{'q'}$ at centre.
The charge $q$ on a capacitor varies with voltage as shown in figure. The area of the triangle $AOB $ represents
What can be said for electric charge if electric flux assocaited with closed loop is zero ?