Question

Let's make some assumptions about numbers to enter in this equation. the visible light output (in w) of a 60 w lightbulb operating at 7% efficiency is approximately 4.2 correct: your answer is correct. w by visible light. (the remaining energy transfers out of the lightbulb by conduction and invisible radiation.) a reasonable distance from the lightbulb to the page might be 0.39 m. substitute these values to find emax (in v/m): emax

What is the value of Emax (in V/m) for the given lightbulb scenario?
A) 4.2 V/m
B) 8.5 V/m
C) 2.1 V/m
D) 6.7 V/m

Answer 1

The value of Emax (in V/m) for the given lightbulb scenario is approximately 6.6 * 10^(-4) V/m.

Step-by-step explanation:

To find the value of Emax (in V/m) for the given lightbulb scenario, we need to substitute the given values into the formula for electric field intensity (E). The formula is E = sqrt((2 * power * efficiency) / (c * distance)), where power is the visible light output of the lightbulb, efficiency is the efficiency of the lightbulb, c is the speed of light, and distance is the distance between the lightbulb and the page.

Substituting the given values:

E = sqrt((2 * 4.2 * 0.07) / (3 * 10^8 * 0.39))

E = sqrt(0.00432 * 10^(-6))

E = 6.6 * 10^(-4) V/m

Therefore, the value of Emax (in V/m) for the given lightbulb scenario is approximately 6.6 * 10^(-4) V/m.