A sphere of radius R carries charge such that its volume charge density is proportional to the square of the distance from the centre. What is the ratio of the magnitude of the electric field at a distance 2 R from the centre to the magnitude of the electric field at a distance of `R//2` from the centre?

Options

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

Explanation

To solve the problem step by step, we will follow the logic presented in the video transcript, while ensuring clarity in each step.

### Step 1: Understand the Charge Density
The volume charge density \( \rho \) is given to be proportional to the square of the distance from the center of the sphere. We can express this as:
\[ \rho = k r^2 \]
where \( k \) is a constant and \( r \) is the distance from the center.

### Step 2: Calculate the Charge Enclosed
To find the electric field using Gauss's law, we need to calculate the charge enclosed within a radius \( r \). The charge enclosed \( Q_{\text{enclosed}} \) can be calculated using the integral:
\[ Q_{\text{enclosed}} = \int_0^r \rho \, dV \]
where \( dV = 4\pi r'^2 dr' \). Substituting \( \rho \):
\[ Q_{\text{enclosed}} = \int_0^r k (r')^2 (4\pi (r')^2) dr' = 4\pi k \int_0^r (r')^4 dr' \]
Calculating the integral:
\[ \int_0^r (r')^4 dr' = \frac{r^5}{5} \]
Thus,
\[ Q_{\text{enclosed}} = 4\pi k \cdot \frac{r^5}{5} = \frac{4\pi k r^5}{5} \]

### Step 3: Apply Gauss's Law
According to Gauss's law, the electric field \( E \) at a distance \( r \) from the center is given by:
\[ \oint E \cdot dS = \frac{Q_{\text{enclosed}}}{\epsilon_0} \]
For a spherical surface of radius \( r \):
\[ E \cdot 4\pi r^2 = \frac{4\pi k r^5}{5\epsilon_0} \]
Thus,
\[ E = \frac{k r^3}{5 \epsilon_0} \]

### Step 4: Electric Field Outside the Sphere
For distances greater than the radius \( R \) of the sphere, the charge enclosed becomes constant:
\[ Q_{\text{enclosed}} = \frac{4\pi k R^5}{5} \]
Using Gauss's law again:
\[ E \cdot 4\pi r^2 = \frac{4\pi k R^5}{5\epsilon_0} \]
Thus,
\[ E = \frac{k R^5}{5 \epsilon_0 r^2} \]

### Step 5: Calculate Electric Field at Specific Distances
1. **At \( r = \frac{R}{2} \)**:
\[ E\left(\frac{R}{2}\right) = \frac{k R^5}{5 \epsilon_0 \left(\frac{R}{2}\right)^2} = \frac{k R^5}{5 \epsilon_0 \cdot \frac{R^2}{4}} = \frac{4k R^3}{5 \epsilon_0} \]

2. **At \( r = 2R \)**:
\[ E(2R) = \frac{k R^5}{5 \epsilon_0 (2R)^2} = \frac{k R^5}{5 \epsilon_0 \cdot 4R^2} = \frac{k R^3}{20 \epsilon_0} \]

### Step 6: Find the Ratio of Electric Fields
Now, we need to find the ratio:
\[ \text{Ratio} = \frac{E(2R)}{E\left(\frac{R}{2}\right)} = \frac{\frac{k R^3}{20 \epsilon_0}}{\frac{4k R^3}{5 \epsilon_0}} = \frac{1}{20} \cdot \frac{5}{4} = \frac{5}{80} = \frac{1}{16} \]

### Final Result
The ratio of the magnitude of the electric field at a distance \( 2R \) from the center to the magnitude of the electric field at a distance \( \frac{R}{2} \) from the center is:
\[ \frac{1}{16} \]