Consider a solid cube made up of insulating material having a uniform volume charge density. Assuming the electrostatic potential to be zero at infinity, the ratio of the potential at a corner of the cube to that at the centre will be

Options

A. `1:1`
B. `1:2`
C. `1:4`
D. `1:8`

Answer

To solve the problem of finding the ratio of the electrostatic potential at a corner of a uniformly charged insulating cube to that at its center, we can follow these steps:

### Step 1: Understanding the Geometry
Consider a solid cube of side length \( a \) with a uniform volume charge density \( \rho \). The cube has 8 corners and a center point. We need to calculate the electrostatic potential at both the center and one of the corners.

### Step 2: Calculate the Potential at the Center
The potential \( V \) due to a volume charge distribution is given by the formula:
\[ V = \frac{1}{4\pi \epsilon_0} \int \frac{\rho \, dV}{r} \]
where \( r \) is the distance from the charge element \( dV \) to the point where we are calculating the potential.

For the center of the cube, the distance from the center to any point inside the cube varies. However, due to symmetry, we can simplify our calculations. The distance from the center of the cube to the center of each face is \( \frac{a}{2} \).

The potential at the center \( V_1 \) can be calculated as:
\[ V_1 = \frac{\rho}{4\pi \epsilon_0} \int \frac{dV}{\left(\frac{a}{2}\right)} = \frac{\rho}{4\pi \epsilon_0} \cdot \frac{8a^3}{\frac{a}{2}} = \frac{16\rho a^2}{4\pi \epsilon_0} \]

### Step 3: Calculate the Potential at a Corner
Now, we calculate the potential at one of the corners of the cube. The distance from the center of the cube to a corner is given by:
\[ r = \sqrt{\left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2 + \left(\frac{a}{2}\right)^2} = \frac{a\sqrt{3}}{2} \]

The potential at the corner \( V_2 \) can be calculated similarly:
\[ V_2 = \frac{\rho}{4\pi \epsilon_0} \int \frac{dV}{\left(\frac{a\sqrt{3}}{2}\right)} = \frac{\rho}{4\pi \epsilon_0} \cdot \frac{8a^3}{\frac{a\sqrt{3}}{2}} = \frac{16\rho a^2}{4\pi \epsilon_0 \sqrt{3}} \]

### Step 4: Calculate the Ratio of Potentials
Now, we find the ratio of the potential at the corner to that at the center:
\[ \frac{V_2}{V_1} = \frac{\frac{16\rho a^2}{4\pi \epsilon_0 \sqrt{3}}}{\frac{16\rho a^2}{4\pi \epsilon_0}} = \frac{1}{\sqrt{3}} \]

### Step 5: Final Result
Thus, the ratio of the potential at a corner of the cube to that at the center is:
\[ \frac{V_2}{V_1} = \frac{1}{\sqrt{3}} \]