The electric flux from a cube of edge l is `phi`. If an edge of the cube is made `2l` and the charge enclosed is halved, its value will be

Options

A. `4phi`
B. `2phi`
C. `phi//2`
D. `phi`

Explanation

To solve the problem step by step, we will use Gauss's Law, which states that the electric flux (Φ) through a closed surface is equal to the charge (Q) enclosed by that surface divided by the permittivity of free space (ε₀).

### Step 1: Understand the initial conditions
Initially, we have a cube with edge length \( l \) and an enclosed charge \( Q \). The electric flux through this cube is given as \( \Phi = \frac{Q}{\epsilon_0} \).

### Step 2: Write the expression for initial flux
From Gauss's Law, we can express the initial electric flux as:
\[ \Phi_1 = \frac{Q}{\epsilon_0} \]
We are given that \( \Phi_1 = \phi \). Therefore:
\[ \phi = \frac{Q}{\epsilon_0} \]

### Step 3: Change the edge length and charge
Now, the edge of the cube is changed to \( 2l \), and the enclosed charge is halved to \( \frac{Q}{2} \).

### Step 4: Write the expression for new flux
Using Gauss's Law again for the new cube with edge length \( 2l \) and charge \( \frac{Q}{2} \):
\[ \Phi_2 = \frac{Q/2}{\epsilon_0} = \frac{Q}{2\epsilon_0} \]

### Step 5: Relate the new flux to the initial flux
We can relate \( \Phi_2 \) to the initial flux \( \phi \):
\[ \Phi_2 = \frac{1}{2} \cdot \frac{Q}{\epsilon_0} = \frac{1}{2} \phi \]

### Step 6: Conclusion
Thus, the new electric flux \( \Phi_2 \) is:
\[ \Phi_2 = \frac{\phi}{2} \]

### Final Answer
The value of the electric flux when the edge of the cube is made \( 2l \) and the charge enclosed is halved is \( \frac{\phi}{2} \).

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