The other two bulbs stay lit with the same brightness.
There is no illustration of the problem provided but I'll attempt to provide an answer.
The relationship between the electric potential difference between two points and the average strength of the electric field between those two points is given by:
║E║ = ΔV/d
║E║ is the magnitude of the average electric field, ΔV is the potential difference between A and B, and d is the distance between A and B.
We are given the following values:
║E║= 10N/C
d = 3m
Plug these values in and solve for ΔV
10 = ΔV/3
ΔV = 30V
Answer: frequency f = 2000 1/s (Hz)
Explanation: period T means time for one cycle and frequency f tells how many cycles there are in seconds. Unit of frequency is 1/s= Hz. Period T = 1/f and frequency f= 1/T.
E.g is T= 0.01 s then f= 1/ (0.01 s) = 100 Hz. And I'd frequency is 20 Hz,
T= 1/(20 Hz)= 0.05 s. So, if T = 0.005 s, f= 1/T
