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

The Correct Option is A

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 \( 2 : 5 \).