If charges q/2 and 2q are placed at the centre of face and at the corner of a cube, then the total flux through the cube will be

Options

A. `(q)/(2epsilon_(@))`
B. `(q)/(epsilon_(@))`
C. `(q)/(6piepsilon_(0))`
D. `(q)/(8epsilon_(@))`

Answer

To find the total electric flux through the cube when charges \( \frac{q}{2} \) and \( 2q \) are placed at the center of a face and at a corner of the cube respectively, we can follow these steps:

### Step 1: Understand the Position of Charges
- The charge \( \frac{q}{2} \) is located at the center of one face of the cube.
- The charge \( 2q \) is located at one of the corners of the cube.

### Step 2: Calculate the Contribution of Charge \( \frac{q}{2} \)
- The total flux \( \Phi \) through a closed surface is given by Gauss's law:
\[ \Phi = \frac{Q_{\text{enc}}}{\epsilon_0} \]
- Since the charge \( \frac{q}{2} \) is entirely within the cube, its contribution to the total flux is:
\[ \Phi_{\frac{q}{2}} = \frac{\frac{q}{2}}{\epsilon_0} = \frac{q}{2\epsilon_0} \]

### Step 3: Calculate the Contribution of Charge \( 2q \)
- The charge \( 2q \) is located at the corner of the cube. Since a cube has 8 corners, only \( \frac{1}{8} \) of this charge is considered to be inside the cube:
\[ Q_{\text{enc, corner}} = \frac{2q}{8} = \frac{q}{4} \]
- The contribution to the total flux from this charge is:
\[ \Phi_{2q} = \frac{\frac{q}{4}}{\epsilon_0} = \frac{q}{4\epsilon_0} \]

### Step 4: Total Flux Calculation
- Now, we can sum the contributions from both charges to find the total flux through the cube:
\[ \Phi_{\text{total}} = \Phi_{\frac{q}{2}} + \Phi_{2q} = \frac{q}{2\epsilon_0} + \frac{q}{4\epsilon_0} \]
- To add these fractions, we need a common denominator:
\[ \Phi_{\text{total}} = \frac{2q}{4\epsilon_0} + \frac{q}{4\epsilon_0} = \frac{3q}{4\epsilon_0} \]

### Final Answer
Thus, the total electric flux through the cube is:
\[ \Phi_{\text{total}} = \frac{3q}{4\epsilon_0} \]