The total flux through the faces of the cube with side of length a ,if a charge `q` is placed at `B` (mid-point of an edge of the cube), is `phi`, and if a charge `q` is placed at `C` (centre of a face of the cube), is `phi_(2)`. Find `(phi_(2))/(phi_(1))`.
'(##CEN_KSR_PHY_JEE_C18_E01_018_Q06##)'

Answer

To find the ratio of the electric flux through the faces of a cube when a charge \( q \) is placed at different positions, we will analyze the two scenarios:

1. When the charge \( q \) is placed at point \( B \) (the midpoint of an edge of the cube).
2. When the charge \( q \) is placed at point \( C \) (the center of a face of the cube).

### Step-by-Step Solution

#### Step 1: Understanding Electric Flux
Electric flux \( \Phi \) through a closed surface is given by Gauss's Law:
\[ \Phi = \frac{q_{\text{enc}}}{\varepsilon_0} \]
where \( q_{\text{enc}} \) is the charge enclosed by the surface and \( \varepsilon_0 \) is the permittivity of free space.

#### Step 2: Flux when charge \( q \) is at point \( B \)
- When the charge \( q \) is placed at point \( B \) (the midpoint of an edge of the cube), it is not enclosed by the cube.
- The charge effectively influences the flux through the cube, but only a fraction of the total flux due to the symmetry of the cube.
- Since the charge is at the midpoint of an edge, it can be considered that the flux through the cube is \( \Phi_1 = \frac{q}{4\varepsilon_0} \).

#### Step 3: Flux when charge \( q \) is at point \( C \)
- When the charge \( q \) is placed at point \( C \) (the center of a face of the cube), the situation changes.
- The charge is effectively enclosed by the cube and contributes to the total flux through the cube.
- The total flux through the cube in this case is \( \Phi_2 = \frac{q}{2\varepsilon_0} \).

#### Step 4: Finding the ratio \( \frac{\Phi_2}{\Phi_1} \)
- Now we can find the ratio of the two fluxes:
\[ \frac{\Phi_2}{\Phi_1} = \frac{\frac{q}{2\varepsilon_0}}{\frac{q}{4\varepsilon_0}} = \frac{q}{2\varepsilon_0} \cdot \frac{4\varepsilon_0}{q} = \frac{4}{2} = 2 \]

### Final Result
Thus, the ratio of the electric flux through the faces of the cube when the charge is at point \( C \) to when it is at point \( B \) is:
\[ \frac{\Phi_2}{\Phi_1} = 2 \]