What is the flux through a cube of side 'a' if a point charge of q is at one of its corner?

Options

A. `q/(2epsilon_0)`
B. `(2q)/epsilon_0`
C. `Q/(8epsilon_0)`
D. `q/epsilon_0`

Answer

To find the electric flux through a cube of side 'a' when a point charge 'q' is located at one of its corners, we can apply Gauss's Law. Here is a step-by-step solution:

### Step 1: Understand Gauss's Law
Gauss's Law states that the electric flux (Φ) through a closed surface is equal to the charge (Q_enc) enclosed by that surface divided by the permittivity of free space (ε₀):
\[ \Phi = \frac{Q_{\text{enc}}}{\varepsilon_0} \]

### Step 2: Analyze the Charge Location
In this case, the charge 'q' is located at one corner of the cube. Since the charge is at the corner, it is not fully enclosed within the cube.

### Step 3: Enclose the Charge
To properly apply Gauss's Law, we can imagine a larger cube that fully encloses the charge 'q'. This larger cube can be visualized as being composed of 8 smaller cubes, each identical to the original cube of side 'a'.

### Step 4: Calculate the Charge Enclosed
Since the charge 'q' is at one corner of the larger cube, it is shared among the 8 smaller cubes. Therefore, the charge enclosed by one of the smaller cubes is:
\[ Q_{\text{enc}} = \frac{q}{8} \]

### Step 5: Apply Gauss's Law
Now, we can substitute the enclosed charge into Gauss's Law:
\[ \Phi = \frac{Q_{\text{enc}}}{\varepsilon_0} = \frac{q/8}{\varepsilon_0} = \frac{q}{8\varepsilon_0} \]

### Step 6: Conclusion
Thus, the electric flux through the cube of side 'a' with the point charge 'q' at one of its corners is:
\[ \Phi = \frac{q}{8\varepsilon_0} \]