Two concentric spherical shells of radii R and 2R are given charges `Q_(1) and Q_(2)` respectively.
The surfaces charge densities of the outer surfaces are equal. Determine the ratio `Q_(1) : Q_(2)`.

Options

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

Explanation

To solve the problem, we need to determine the ratio of the charges \( Q_1 \) and \( Q_2 \) on two concentric spherical shells with radii \( R \) and \( 2R \) respectively, given that the surface charge densities on their outer surfaces are equal.

### Step-by-Step Solution:

1. **Understand Surface Charge Density**:
The surface charge density \( \sigma \) is defined as the charge per unit area on the surface of the sphere. It can be expressed mathematically as:
\[ \sigma = \frac{Q}{A} \]
where \( Q \) is the charge on the surface and \( A \) is the surface area.

2. **Calculate Surface Areas**:
The surface area \( A \) of a sphere is given by the formula:
\[ A = 4\pi r^2 \]
- For the inner shell (radius \( R \)):
\[ A_1 = 4\pi R^2 \]
- For the outer shell (radius \( 2R \)):
\[ A_2 = 4\pi (2R)^2 = 4\pi (4R^2) = 16\pi R^2 \]

3. **Express Surface Charge Densities**:
Using the surface area formulas, we can express the surface charge densities for both shells:
- For the inner shell:
\[ \sigma_1 = \frac{Q_1}{A_1} = \frac{Q_1}{4\pi R^2} \]
- For the outer shell:
\[ \sigma_2 = \frac{Q_2}{A_2} = \frac{Q_2}{16\pi R^2} \]

4. **Set the Surface Charge Densities Equal**:
According to the problem, the surface charge densities are equal:
\[ \sigma_1 = \sigma_2 \]
Substituting the expressions for \( \sigma_1 \) and \( \sigma_2 \):
\[ \frac{Q_1}{4\pi R^2} = \frac{Q_2}{16\pi R^2} \]

5. **Simplify the Equation**:
We can cancel \( 4\pi R^2 \) from both sides:
\[ Q_1 = \frac{Q_2}{4} \]

6. **Determine the Ratio**:
Rearranging gives us:
\[ \frac{Q_1}{Q_2} = \frac{1}{4} \]
Thus, the ratio of the charges is:
\[ Q_1 : Q_2 = 1 : 4 \]

### Final Answer:
The ratio \( Q_1 : Q_2 \) is \( 1 : 4 \).