NEET Gauss's Law MCQs & PYQs

Assertion (A) is true. An external charge does not change the net charge enclosed by the Gaussian surface \(q_{in}\), so the total electric flux \(\oint \vec{E}.d\vec{s}\) remains unchanged as per Gauss's Law. However, the electric field \(\vec{E}\) at any point on the surface is the vector sum of fields from all charges, both inside and outside, so it will change. Reason (R) is false because the electric field at any point is due to both internal and external charges.

Continuity equation describes conservation of charge. Gauss's law is a fundamental law valid irrespective of the inverse square law and does not show 'diversion' based on its obedience. Thus, (A) is true, (R) is false.

According to Gauss's Law, total flux through the cube is \(\phi = \frac{q}{\epsilon_0}\). Since a cube has 6 identical faces, the flux through one face is \(\phi' = \frac{\phi}{6} = \frac{Q \times 10^{-6}}{6\epsilon_0}\).

Gauss's law states that the net electric flux \(\phi\) through any closed surface is equal to \(1/varepsilon_0\) times the net charge enclosed by the surface (\(\phi = q_{\text{enclosed}}/\varepsilon_0\)), independent of the shape, area, or volume of the surface.

According to Gauss's law, the net electric flux through a closed surface is given by \(\Phi = \frac{q_{\text{encl}}}{\varepsilon_0}\), which only depends on the total charge enclosed.

The electric fields due to both line charges at the mid-point point in the same direction. Magnitude of field due to each line is $E = \frac{\lambda}{2\pi\epsilon_0 R}$. Total field $E_{net} = E + E = \frac{\lambda}{\pi\epsilon_0 R}$.

According to Gauss's law, the electric flux $\Phi = \oint E \cdot dA = \frac{q_{in}}{\epsilon_0}$. The electric field at radius $a$ is $E = Aa$. Therefore, flux $\Phi = (Aa)(4\pi a^2) = 4\pi A a^3$. Equating this to $\frac{q_{in}}{\epsilon_0}$, we get $q_{in} = 4\pi\epsilon_0 A a^3$.

According to Gauss's law, the total outward electric flux through any closed surface depends only on the net charge enclosed ($Q$). Since the enclosed charge remains the same when the radius is doubled, the flux will remain the same.

Total electric flux through the entire hollow cylinder is $\Phi_{total} = \frac{q}{\epsilon_0}$. This total flux is the sum of fluxes through the two plane surfaces (A and C) and the curved surface (B). By symmetry, $\Phi_A = \Phi_C$. Thus, $2\Phi_A + \phi = \frac{q}{\epsilon_0}$, which gives $\Phi_A = \frac{1}{2}(\frac{q}{\epsilon_0} - \phi)$.

The electric field at a point on the equatorial plane of a dipole is opposite to the dipole moment direction. Its magnitude is $E = \frac{p}{4\pi\epsilon_0 r^3}$. Thus in vector form, $\vec{E} = -\frac{\vec{p}}{4\pi \epsilon_0 r^3}$.