- Home
- Standard 12
- Physics
A point charge causes an electric flux of $-1.0 \times 10^{3}\; N\;m ^{2} / C$ to pass through a spherical Gaussian surface of $10.0\; cm$ radius centred on the charge.
$(a)$ If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
$(b)$ What is the value of the point charge?
Solution
$(a)$ Electric flux, $\phi=-1.0 \times 10^{3}\, N\, m ^{2} / C$ and radius of the Gaussian surface, $r =10.0 \;cm$ Electric flux piercing out through a surface depends on the net charge enclosed inside a body.
It does not depend on the size of the body. If the radius of the Gaussian surface is doubled, then the flux passing through the surface remains the same i.e., $-10^{3}\; N\; m ^{2} / C$
$(b)$ Electric flux is given by the relation $\quad \phi=\frac{q}{\varepsilon_{0}}$
Where, $\varepsilon_{0}=$ Permittivity of free space $=8.854 \times 10^{-12}\, N ^{-1} \,C ^{2}\, m ^{-2}$
$q =$ Net charge enclosed by the spherical surface $=\phi \varepsilon_{0}$ $=-1.0 \times 10^{3} \times 8.854 \times 10^{-12}=-8.854 \times 10^{-9} \,C$$=-8.854 \,n\,C$
Therefore, the value of the point charge is $-8.854 \,n\,C$.
