The electric field intensity produced by the radiations coming from 100W bulbs at a 3m distance is E. The electric field intensity produced by the radiations coming from 50W bulb at the same distance is

Options

A. `E/2`
B. `2E`
C. `E/(sqrt(2))`
D. `sqrt(2)E`

Explanation

To solve the problem, we need to determine the electric field intensity produced by a 50W bulb at a distance of 3m, given that the electric field intensity produced by a 100W bulb at the same distance is E.

### Step-by-Step Solution:

1. **Understand the relationship between electric field intensity and power**:
The electric field intensity (E) produced by a source is directly proportional to the power (P) of that source. This can be expressed as:
\[ E \propto P \]

2. **Set up the proportionality for the two bulbs**:
Let \( E_1 \) be the electric field intensity from the 100W bulb, and \( E_2 \) be the electric field intensity from the 50W bulb. We can express this relationship as:
\[ \frac{E_2}{E_1} = \frac{P_2}{P_1} \]
where \( P_1 = 100 \, \text{W} \) and \( P_2 = 50 \, \text{W} \).

3. **Substitute the known values**:
We know that \( E_1 = E \), \( P_1 = 100 \, \text{W} \), and \( P_2 = 50 \, \text{W} \). Therefore, we can write:
\[ \frac{E_2}{E} = \frac{50}{100} \]

4. **Simplify the equation**:
Simplifying the right side gives:
\[ \frac{E_2}{E} = \frac{1}{2} \]

5. **Solve for \( E_2 \)**:
To find \( E_2 \), we multiply both sides by \( E \):
\[ E_2 = \frac{1}{2} E \]

6. **Conclusion**:
Thus, the electric field intensity produced by the radiations coming from a 50W bulb at a distance of 3m is:
\[ E_2 = \frac{E}{2} \]

### Final Answer:
The electric field intensity produced by the radiations coming from a 50W bulb at the same distance is \( \frac{E}{2} \).

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