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 of finding the ratio of charge density on the surfaces of two isolated metallic spheres of radii 2 cm and 4 cm that are given equal charge, we can follow these steps:

### Step 1: Define the Variables
Let:
- Radius of the first sphere, \( R_1 = 2 \, \text{cm} = 0.02 \, \text{m} \)
- Radius of the second sphere, \( R_2 = 4 \, \text{cm} = 0.04 \, \text{m} \)
- Charge on both spheres, \( Q_1 = Q_2 = Q \)

### Step 2: Calculate the Surface Area of Each Sphere
The surface area \( A \) of a sphere is given by the formula:
\[ A = 4\pi R^2 \]
For the first sphere:
\[ A_1 = 4\pi R_1^2 = 4\pi (0.02)^2 = 4\pi (0.0004) = 0.0048\pi \, \text{m}^2 \]
For the second sphere:
\[ A_2 = 4\pi R_2^2 = 4\pi (0.04)^2 = 4\pi (0.0016) = 0.0064\pi \, \text{m}^2 \]

### Step 3: Calculate the Surface Charge Density
The surface charge density \( \sigma \) is defined as the charge per unit area:
\[ \sigma = \frac{Q}{A} \]
For the first sphere:
\[ \sigma_1 = \frac{Q}{A_1} = \frac{Q}{0.0048\pi} \]
For the second sphere:
\[ \sigma_2 = \frac{Q}{A_2} = \frac{Q}{0.0064\pi} \]

### Step 4: Find the Ratio of Charge Densities
Now, we can find the ratio of the surface charge densities:
\[ \frac{\sigma_1}{\sigma_2} = \frac{\frac{Q}{0.0048\pi}}{\frac{Q}{0.0064\pi}} = \frac{0.0064\pi}{0.0048\pi} \]
The \( Q \) and \( \pi \) cancel out:
\[ \frac{\sigma_1}{\sigma_2} = \frac{0.0064}{0.0048} = \frac{6.4}{4.8} = \frac{4}{3} \]

### Step 5: Simplify the Ratio
To simplify:
\[ \frac{4}{3} = \frac{4}{1} \]

### Conclusion
Thus, the ratio of charge density on the surfaces of the spheres is:
\[ \frac{\sigma_1}{\sigma_2} = 4 : 1 \]