Question

(2 points) A pair of students measure the intensity of light from a desklamp in Sec. 4.5 of the experiment. When the detector is 25 cm from the lamp, the signal measured by the detector is 0.74 V. How large a signal (in V) do the students expect when the detector is 47 cm away?

Answer 1

Answer:

At 47 cm signal will be of 0.2093 volt

Explanation:

We have given the the the signal measured by the detector at a distance of 25 cm is 0.74 volt

We have to find the intensity of the signal at a distance of 47 cm

We know that intensity inversely proportional to the square of distance

So $\frac{I_1}{I_2}=\frac{R_2^2}{R_1^2}$, here $I_1$  is intensity at 25 cm and  $I_2$ is intensity at 47 cm

So $\frac{0.74}{I_2}=\frac{47^2}{25^2}$

$I_2=0.2093volt$

Answer 2

This question involves the concept of electric field intensity and distance from the source.

The students expect a signal of "0.21 V".

The formula for the electric field intensity is given as follows:

$E = \frac{Q}{4\pi \epsilon_oR^2}$

where,

E = Electric field intensity

Q = Electric Charge

R = Distance from source

Therefore, it is clear that the intensity is inversely proportional to the square of the distance.

$E\ \alpha\ \frac{1}{R^2}\\\\ER^2=constant$

Hence,

$E_1R_1^2=E_2R_2^2$

where,

E₁ = Initial Intensity = 0.74 V

R₁ = Initial Distance = 25 cm

E₂ = Final Intensity = ?

R₂ = Final Distance = 47 cm

Therefore,

$(0.74\ V)(25\ cm)^2=(E_2)(47\ cm)^2\\\\E_2 = \frac{(0.74\ V)(25\ cm)^2}{(47\ cm)^2}$

E₂ = 0.21 V

Learn more about electric field intensity here:

https://brainly.com/question/23522152?referrer=searchResults

The attached picture shows electric field intensity.