Two isolated metallic spheres of radii 2 cm and 4 cm are given equal charge, then the ratio of charge density on the surfaces of the spheres will be

Options

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

Answer

To solve the problem, we need to find the ratio of the charge densities on the surfaces of two isolated metallic spheres with given radii. Let's break down the steps:

### Step 1: Define the Problem
We have two metallic spheres:
- Sphere 1 with radius \( r_1 = 2 \, \text{cm} \)
- Sphere 2 with radius \( r_2 = 4 \, \text{cm} \)

Both spheres are given equal charge \( Q \).

### Step 2: Understand Surface Charge Density
The surface charge density \( \sigma \) is defined as the charge \( Q \) divided by the surface area \( A \) of the sphere:
\[ \sigma = \frac{Q}{A} \]
For a sphere, the surface area \( A \) is given by:
\[ A = 4\pi r^2 \]

### Step 3: Calculate Surface Charge Densities
For Sphere 1:
\[ \sigma_1 = \frac{Q}{4\pi r_1^2} = \frac{Q}{4\pi (2 \, \text{cm})^2} = \frac{Q}{4\pi \cdot 4 \, \text{cm}^2} = \frac{Q}{16\pi \, \text{cm}^2} \]

For Sphere 2:
\[ \sigma_2 = \frac{Q}{4\pi r_2^2} = \frac{Q}{4\pi (4 \, \text{cm})^2} = \frac{Q}{4\pi \cdot 16 \, \text{cm}^2} = \frac{Q}{64\pi \, \text{cm}^2} \]

### Step 4: Find the Ratio of Charge Densities
Now, we need to find the ratio \( \frac{\sigma_1}{\sigma_2} \):
\[ \frac{\sigma_1}{\sigma_2} = \frac{\frac{Q}{16\pi \, \text{cm}^2}}{\frac{Q}{64\pi \, \text{cm}^2}} = \frac{Q}{16\pi} \times \frac{64\pi}{Q} \]
The \( Q \) and \( \pi \) cancel out:
\[ \frac{\sigma_1}{\sigma_2} = \frac{64}{16} = 4 \]

### Step 5: Conclusion
Thus, the ratio of the charge densities on the surfaces of the spheres is:
\[ \sigma_1 : \sigma_2 = 4 : 1 \]

### Final Answer
The ratio of charge density on the surfaces of the spheres is \( 4 : 1 \).

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