Two spheres `A` and `B` of radius 'a' and 'b' respectively are at same electric potential. The ratio of the surface charge densities of `A` and `B` is

Options

A. `(a)/(b)`
B. `(b)/(a)`
C. `(a^(2))/(b^(2))`
D. `(b^(2))/(a^(2))`

Answer

To solve the problem of finding the ratio of the surface charge densities of two spheres A and B that are at the same electric potential, we can follow these steps:

### Step 1: Understand the Electric Potential of a Sphere
The electric potential \( V \) of a charged sphere with radius \( r \) and total charge \( Q \) is given by the formula:
\[ V = \frac{KQ}{r} \]
where \( K \) is the electrostatic constant.

### Step 2: Relate Charge to Surface Charge Density
The surface charge density \( \sigma \) is defined as the charge per unit area. For a sphere, the total charge \( Q \) can be expressed as:
\[ Q = \sigma \times \text{Area} = \sigma \times 4\pi r^2 \]
Thus, substituting this into the potential formula gives:
\[ V = \frac{K(\sigma \times 4\pi r^2)}{r} = \frac{4\pi K \sigma r}{1} \]

### Step 3: Express Potential in Terms of Surface Charge Density
For sphere A with radius \( a \) and surface charge density \( \sigma_A \):
\[ V_A = \frac{4\pi K \sigma_A a}{1} \]
For sphere B with radius \( b \) and surface charge density \( \sigma_B \):
\[ V_B = \frac{4\pi K \sigma_B b}{1} \]

### Step 4: Set the Potentials Equal
Since both spheres are at the same electric potential:
\[ V_A = V_B \]
This leads to:
\[ \frac{4\pi K \sigma_A a}{1} = \frac{4\pi K \sigma_B b}{1} \]

### Step 5: Simplify the Equation
We can cancel \( 4\pi K \) from both sides:
\[ \sigma_A a = \sigma_B b \]

### Step 6: Find the Ratio of Surface Charge Densities
Rearranging the equation gives:
\[ \frac{\sigma_A}{\sigma_B} = \frac{b}{a} \]

### Conclusion
Thus, the ratio of the surface charge densities of spheres A and B is:
\[ \frac{\sigma_A}{\sigma_B} = \frac{b}{a} \]

### Final Answer
The ratio of the surface charge densities of spheres A and B is \( \frac{b}{a} \).

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