Question:
The electric field intensity produced by the radiations coming from 100W bulbs at 3m distance is E. The electric field intensity produced by the radiations coming from 50W bulbs at the same distance is:
The electric field intensity produced by the radiations coming from 100W bulbs at 3m distance is E. The electric field intensity produced by the radiations coming from 50W bulbs at the same distance is:
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When radiation power is scaled, remember that the electric field magnitude scales with the square root of power, while the intensity scales linearly with power.
Updated On: Oct 7, 2026
- \( E/2 \)
- \( 2E \)
- \( E/\sqrt{2} \)
- \( \sqrt{2}E \)
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The Correct Option is C
Solution and Explanation
Concept:
The intensity \( I \) of electromagnetic radiation from a point source is defined as the power per unit area, \( I = \frac{P}{4\pi r^2} \). The intensity is also related to the electric field amplitude \( E_0 \) by the relation \( I = \frac{1}{2} c \epsilon_0 E_0^2 \). From these two equations, it follows that the amplitude of the electric field \( E_0 \) is proportional to the square root of the power \( P \), as \( E_0^2 \propto P \implies E_0 \propto \sqrt{P} \).
Step 1: Establish the proportional relationship.
Since the distance \( r \) is kept constant for both bulbs, the ratio of the electric field intensities \( E_1 \) and \( E_2 \) depends solely on the power ratio of the bulbs: $$ \frac{E_1}{E_2} = \sqrt{\frac{P_1}{P_2}} $$
Step 2: Substitute the given power values.
Given \( P_1 = 100 \text{ W} \), \( P_2 = 50 \text{ W} \), and \( E_1 = E \): $$ \frac{E}{E_2} = \sqrt{\frac{100}{50}} $$ $$ \frac{E}{E_2} = \sqrt{2} $$
Step 3: Solve for \( E_2 \).
$$ E_2 = \frac{E}{\sqrt{2}} $$ This indicates that reducing the power of the source by half results in the electric field amplitude decreasing by a factor of \( \sqrt{2} \). $$\boxed{E/\sqrt{2}}$$
Step 1: Establish the proportional relationship.
Since the distance \( r \) is kept constant for both bulbs, the ratio of the electric field intensities \( E_1 \) and \( E_2 \) depends solely on the power ratio of the bulbs: $$ \frac{E_1}{E_2} = \sqrt{\frac{P_1}{P_2}} $$
Step 2: Substitute the given power values.
Given \( P_1 = 100 \text{ W} \), \( P_2 = 50 \text{ W} \), and \( E_1 = E \): $$ \frac{E}{E_2} = \sqrt{\frac{100}{50}} $$ $$ \frac{E}{E_2} = \sqrt{2} $$
Step 3: Solve for \( E_2 \).
$$ E_2 = \frac{E}{\sqrt{2}} $$ This indicates that reducing the power of the source by half results in the electric field amplitude decreasing by a factor of \( \sqrt{2} \). $$\boxed{E/\sqrt{2}}$$
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