Learn gauss's law with examples, practice to calculate flux linked with any face of cube when charge is placed at center and the flux linked with a spherical surface length of a uniformly charged rod.
\(\phi_E=\int\limits_s\vec E.d\vec a\)
\(\phi\propto q_{inside}\)
or, \(\phi=\dfrac {q_{inside}}{\epsilon_0}\)
where, \(\epsilon_0\) is the permittivity of the free space.
\(\phi=\oint\limits_s \vec E.d\vec a=\dfrac {Q_{inside}}{\epsilon_0}\)
\(\phi\propto q_{inside}\)
or, \(\phi=\dfrac {q_{inside}}{\mathcal E_0}\)
where, \(\mathcal E_0\) is the permittivity of the free space.
\(\phi=\oint\limits_s \vec E.d\vec a=\dfrac {Q_{inside}}{\mathcal E_0}\)
A 0 Nm2/C
B \(\dfrac {18}{\epsilon_0}\) Nm2/C
C \(\dfrac {12}{\epsilon_0}\) Nm2/C
D \(\dfrac {6}{\epsilon_0}\) Nm2/C
\(\phi=\dfrac {{\text {Total charge enclosed by spherical surface}}} {\epsilon_0}\)
\(\phi=\dfrac {\text{(Charge per unit length of rod × length of rod enclosed by sphere)+all point charges inside}} {\epsilon_0}\)
A \(300\epsilon_0\) Nm2/C
B \(\dfrac {11\epsilon_0}{3}\) Nm2/C
C \(110\epsilon_0\) Nm2/C
D \(\dfrac {11}{3\epsilon_0}\) Nm2/C
So, net flux through any face of cube will be same and is given as \(=\dfrac {q}{6\epsilon_0}\)
A \(\dfrac{q}{2\epsilon_0}\)
B 0
C \(\dfrac{q}{6\epsilon_0}\)
D \(\dfrac{q}{4\epsilon_0}\)
Then charge enclosed by closed surface is given as: -
Charge enclosed by closed surface = Length of rod enclosed by surface × Charge per unit length
A \(\dfrac {100}{\epsilon_0}\)Nm2/C
B \(\dfrac {2.5}{\epsilon_0}\)Nm2/C
C \(2.5\epsilon_0\)Nm2/C
D \(2.5\epsilon_0^2\) Nm2/C
So, charge per unit Area = \(\dfrac {Total\;charge}{Total\;Area}\) = \(\dfrac {Q}{A }\)
Charge enclosed by the surface = Area of sheet enclosed by Gaussian surface(A') × charge per unit area\(\left( \dfrac {Q}{A} \right)\)
Flux linked with Gaussian surface = \(\dfrac {Total\;charge\;enclosed\;by\;the\;surface}{\epsilon_0}\)
A \({3}{\epsilon_0}\) Nm2/C
B \(\dfrac{4}{5\epsilon_0}\) Nm2/C
C \(\dfrac{4}{\epsilon_0}\) Nm2/C
D \(\dfrac{3}{\epsilon_0}\)Nm2/C