Question 777: Electrostats & Capacitance

Question

A charge $\mathrm{Q} \mu \mathrm{C}$ is placed at the centre of a cube. The flux through two opposite faces of the cube is ( $\varepsilon_0=$ permittivity of free space)

Accepted Answer

Step 1: Recall GaussÃ¢ÂÂs law
$ \Phi_{	ext{total}} = \frac{Q}{\varepsilon_0} $
This is the total electric flux through a closed surface enclosing charge $Q$.
Here the closed surface is a cube around the charge.
Step 2: Symmetry
Since the charge is exactly at the cubeÃ¢ÂÂs centre, by symmetry, the electric flux will be evenly distributed among all six faces of the cube.
So, flux through one face =
$ \Phi_{	ext{one face}} = \frac{Q}{6 \varepsilon_0} $
Step 3: Flux through two opposite faces
Flux through two faces =
$ 2 	imes \frac{Q}{6 \varepsilon_0} = \frac{Q}{3 \varepsilon_0} $
Final Answer:
$ \boxed{\frac{Q}{3\varepsilon_0}} $
Ã°ÂÂÂ Correct option:
B

Answer

$\frac{\mathrm{Q}}{6 \varepsilon_0}$

Answer

$\frac{\mathrm{Q}}{3 \varepsilon_0}$

Answer

$\frac{\mathrm{Q}}{\varepsilon_0}$

Answer

$\frac{\mathrm{Q}}{2 \varepsilon_0}$

Electrostats & Capacitance

A charge \mathrm{Q} \mu \mathrm{C} is placed at the centre of a cube. The flux through two opposite faces of the cube is ( \varepsilon_0= permittivity of free space)

Choose the correct answer: