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:

• A. \( E/2 \)
• B. \( 2E \)
• C. \( E/\sqrt{2} \)
• D. \( \sqrt{2}E \)

Hint:

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.

Answer 1

The Correct Option is C

Step 1: Intensity from a bulb.
Treating the bulb as a point source, the intensity at distance $r$ is $I = \dfrac{P}{4\pi r^2}$, the power spread over a sphere.
Step 2: Intensity and electric field.
For an electromagnetic wave, $I = \dfrac{1}{2}c\varepsilon_0 E_0^2$, so $E_0^2 \propto I$.
Step 3: Combine the two.
Putting them together, $E_0^2 \propto \dfrac{P}{r^2}$, so $E_0 \propto \dfrac{\sqrt{P}}{r}$. At a fixed distance, $E_0 \propto \sqrt{P}$.
Step 4: Form the ratio.
Since $r$ is the same $3\,\text{m}$ for both bulbs, \[ \frac{E_1}{E_2} = \sqrt{\frac{P_1}{P_2}}. \]
Step 5: Put in the powers.
With $P_1 = 100\,\text{W}$, $P_2 = 50\,\text{W}$, and $E_1 = E$: \[ \frac{E}{E_2} = \sqrt{\frac{100}{50}} = \sqrt{2}. \]
Step 6: Solve for the second field.
$E_2 = \dfrac{E}{\sqrt{2}}$.
\[ \boxed{E/\sqrt{2}} \]