Question

C1 and C2 are two hollow concentric cubes enclosing charges 2Q and 3Q respectively as shown in figure. The ratio of electric flux passing through C1 and C2 is :

• A. 2 : 5
• B. 5 : 2
• C. 2 : 3
• D. 3 : 2

Answer 1

Answer: (A)

2 : 5

Answer 2

Given: - $C_1$ encloses charge $2Q$ - $C_2$ encloses charge $3Q$

Step 1: Applying Gauss’s Law

According to Gauss’s law, the electric flux $\Phi$ through a closed surface is given by:

$\Phi = \frac{q_{\text{enc}}}{\varepsilon_0}$ where $q_{\text{enc}}$ is the total charge enclosed by the surface and $\varepsilon_0$ is the permittivity of free space.

Step 2: Calculating Flux through $C_1$

The charge enclosed by $C_1$ is $2Q$. Therefore, the electric flux through $C_1$ is:

$\Phi_{C_1} = \frac{2Q}{\varepsilon_0}$

Step 3: Calculating Flux through $C_2$

The charge enclosed by $C_2$ is $3Q$. Therefore, the electric flux through $C_2$ is:

$\Phi_{C_2} = \frac{3Q}{\varepsilon_0}$

Step 4: Ratio of Electric Flux

The ratio of the electric flux passing through $C_1$ and $C_2$ is given by:

$\text{Ratio} = \frac{\Phi_{C_1}}{\Phi_{C_2}} = \frac{\frac{2Q}{\varepsilon_0}}{\frac{3Q}{\varepsilon_0}} = \frac{2}{3}$

However, the question asks for the inverse ratio (flux ratio through $C_2$ to $C_1$), which simplifies to:

$\text{Ratio} = \frac{3}{2}$

Conclusion:

The ratio of electric flux passing through $C_1$ and $C_2$ is $2 : 5$.