Gauss’s Law Dependency – Rankers Physics

Gauss's Law: Practice Problem & Solution

According to Gauss law of electrostatics, electric flux through a closed surface depends on
The shape of the surface
The volume enclosed by the surface
The area of the surface
The quantity of charges enclosed by the surface

Solution Explained:

To solve this problem, we apply the core principles of Gauss's Law. Understanding the underlying formula is key to arriving at the correct answer below:

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.

Leave a Reply

Your email address will not be published. Required fields are marked *

Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. Three point charges $+q$, $-2q$ and $+q$ are placed at points $(x=0, y=a, z=0)$, $(x=0, y=0, z=0)$ and $(x=a, y=0, z=0)$ respectively. The magnitude and direction of the electric dipole moment vector of this charge assembly are: (2007)
2. In the arrangement of charges shown, the dotted boundary represents a Gaussian surface . For this surface , Gauss's law has the form \[ \oint_{}^{}\overrightarrow{E}.\overrightarrow{ds}=\frac{q}{\varepsilon_{0}} .\] Select the correct alternative \[ \overrightarrow{E_{1}},\overrightarrow{E_{2}}, \overrightarrow{E_{3}} and \overrightarrow{E_{4}} \] are electric field vectors due to these four charges respectively.
3. If some positive charge is given to a solid conductor, then its potential is
4. Assertion (A): If electric field in x-y plane is given by \( \vec{E} = y \hat{i} + x \hat{j} \) then equipotential curve is given by \( xy = \text{constant} \). Reason (R): Electric field may not be perpendicular to equipotential surface/curve/line.  
5. Assertion (A): A metal sphere of radius \(1\text{ cm}\) cannot hold a charge of \(1\text{ coulomb}\) in air. Reason (R): The dielectric strength of air (minimum field required for ionisation of a medium) is \(3\text{ MV/m}\).