Two charged spheres of radii `R_(1) and R_(2)` having equal surface charge density. The ratio of their potential is

Options

A. `R_(1)//R_(2)`
B. `R_(2)//R_(1)`
C. `(R_(1)//R_(2))^(2)`
D. `(R_(2)//R_(1))^(2)`

Answer

To find the ratio of the potentials of two charged spheres with equal surface charge densities, we can follow these steps:

### Step 1: Understand the relationship between charge, surface charge density, and area
The surface charge density (\( \sigma \)) is defined as the charge (\( Q \)) per unit area (\( A \)). The area of a sphere is given by \( A = 4\pi R^2 \). Therefore, the charge on each sphere can be expressed as:
\[ Q_1 = \sigma \cdot A_1 = \sigma \cdot 4\pi R_1^2 \]
\[ Q_2 = \sigma \cdot A_2 = \sigma \cdot 4\pi R_2^2 \]

### Step 2: Write the expression for the electric potential of a charged sphere
The electric potential (\( V \)) at the surface of a charged sphere is given by the formula:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant.

### Step 3: Substitute the expressions for charge into the potential formula
For sphere 1:
\[ V_1 = \frac{kQ_1}{R_1} = \frac{k(\sigma \cdot 4\pi R_1^2)}{R_1} \]
This simplifies to:
\[ V_1 = k \sigma \cdot 4\pi R_1 \]

For sphere 2:
\[ V_2 = \frac{kQ_2}{R_2} = \frac{k(\sigma \cdot 4\pi R_2^2)}{R_2} \]
This simplifies to:
\[ V_2 = k \sigma \cdot 4\pi R_2 \]

### Step 4: Find the ratio of the potentials
Now we can find the ratio of the potentials \( V_1 \) and \( V_2 \):
\[ \frac{V_1}{V_2} = \frac{k \sigma \cdot 4\pi R_1}{k \sigma \cdot 4\pi R_2} \]
The constants \( k \), \( \sigma \), and \( 4\pi \) cancel out:
\[ \frac{V_1}{V_2} = \frac{R_1}{R_2} \]

### Conclusion
Thus, the ratio of the potentials of the two charged spheres is:
\[ \frac{V_1}{V_2} = \frac{R_1}{R_2} \]

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