The electric field intensity at a distance `20`cm from the centre of a uniformly charged non conducting solid sphere of radius `10`cm is `E`.Then what is the electric field intensity at a distance `5`cm from the `centre it will be.....

Options

A. 16 E
B. 4 E
C. 2E
D. Zero

Answer

To solve the problem, we need to find the electric field intensity at a distance of 5 cm from the center of a uniformly charged non-conducting solid sphere, given that the electric field intensity at a distance of 20 cm from the center is E.

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Radius of the sphere (R) = 10 cm
- Distance from the center for the first case (r) = 20 cm
- Distance from the center for the second case (r') = 5 cm
- Electric field intensity at 20 cm = E

2. **Determine the Region**:
- Since r (20 cm) is greater than R (10 cm), we are outside the sphere for the first case.
- For the second case, since r' (5 cm) is less than R (10 cm), we are inside the sphere.

3. **Electric Field Outside the Sphere**:
- The electric field intensity (E) outside a uniformly charged non-conducting solid sphere is given by:
\[ E = \frac{kQ}{r^2} \]
- Here, \( k \) is Coulomb's constant, \( Q \) is the total charge, and \( r \) is the distance from the center.

4. **Electric Field Inside the Sphere**:
- The electric field intensity (E') inside a uniformly charged non-conducting solid sphere is given by:
\[ E' = \frac{kQ}{R^3} r' \]
- Here, \( R \) is the radius of the sphere and \( r' \) is the distance from the center.

5. **Relate the Two Electric Fields**:
- From the first case, we have:
\[ E = \frac{kQ}{(20)^2} \]
- From the second case, we have:
\[ E' = \frac{kQ}{(10)^3} (5) \]

6. **Divide the Two Equations**:
- To find the relationship between E' and E, we can divide the equation for E' by the equation for E:
\[ \frac{E'}{E} = \frac{\frac{kQ}{(10)^3} (5)}{\frac{kQ}{(20)^2}} \]

7. **Simplify the Equation**:
- The \( kQ \) terms cancel out:
\[ \frac{E'}{E} = \frac{(5)(20^2)}{(10^3)} \]
- Simplifying further:
\[ \frac{E'}{E} = \frac{5 \cdot 400}{1000} = \frac{2000}{1000} = 2 \]

8. **Final Result**:
- Therefore, we find:
\[ E' = 2E \]
- The electric field intensity at a distance of 5 cm from the center of the sphere is \( 2E \).