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

The problem requires finding the ratio of electric flux passing through two concentric hollow cubes, C1 and C2. We can solve this using Gauss's Law, which states that the electric flux \(\Phi\) through a closed surface is proportional to the charge enclosed by that surface. The formula is given by:

\(\Phi = \frac{Q_{\text{enclosed}}}{\varepsilon_0}\)

Calculate the electric flux through cube C1:

Since cube C1 encloses a charge of \(2Q\), the electric flux through C1, \(\Phi_1\), is:

\(\Phi_1 = \frac{2Q}{\varepsilon_0}\)

Calculate the electric flux through cube C2:

Cube C2 encloses both the charge inside C1 and its own charge, totaling \(2Q + 3Q = 5Q\). Therefore, the electric flux through C2, \(\Phi_2\), is:

\(\Phi_2 = \frac{5Q}{\varepsilon_0}\)

Find the ratio of electric flux through C1 and C2:

The ratio \(\frac{\Phi_1}{\Phi_2}\) is:

\(\frac{\Phi_1}{\Phi_2} = \frac{\frac{2Q}{\varepsilon_0}}{\frac{5Q}{\varepsilon_0}} = \frac{2}{5}\)

Hence, the ratio of electric flux passing through C1 and C2 is \(2 : 5\).

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 \( 2 : 5 \).