Rankers Physics

Gauss's Law: Practice Problem & Solution

A charge $Q \mu\text{C}$ is placed at the centre of a cube, the flux coming out from any surface will be: (2001)
$\frac{Q \times 10^{-6}}{6\varepsilon_0}$
$\frac{Q \times 10^{-3}}{6\varepsilon_0}$
$\frac{Q}{24\varepsilon_0}$
$\frac{Q}{8\varepsilon_0}$

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:

The enclosed charge is $Q \mu\text{C} = Q \times 10^{-6} \text{ C}$. By Gauss's Law, total flux through the cube is $\frac{Q \times 10^{-6}}{\varepsilon_0}$. Since a cube has 6 symmetric faces, flux through any one surface is $\frac{Q \times 10^{-6}}{6\varepsilon_0}$.

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Your email address will not be published.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. Two point charges exert a force \(F_0\) on each other when placed in vacuum. Now the charges are increased to four times, separation between them is doubled and the system is placed is an insulating medium. Now they experience the same force. What should be the dielectric constant of the medium?
2. 8. A charge $q$ is placed at the centre of the line joining two exactly equal positive charges $Q$. The system of three charges will be in equilibrium, if $q$ is equal to (1995)
3. An electric dipole is placed at an angle of $30^\circ$ with an electric field intensity $2 \times 10^5 \text{ N/C}$. It experiences a torque equal to $4 \text{ Nm}$. The charge on the dipole, if the dipole length is $2 \text{ cm}$, is: (2016-II)
4. The electrostatic potential on the surface of a charged solid conducting sphere is \( 100 \text{ volts} \). Two statements are made in this regard : Assertion (A): At any point inside the sphere, electrostatic potential is \( 100 \text{ volt} \). Reason (R): At any point inside the sphere, electric field is zero.  
5. 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.