Answer:
A) 8.03Hz
Explanation:
f = V/λ
Where wavelength( λ )= 30m
Speed (V) =241m/S
f= 241/30=8.03Hz
Missing question in the text:
"A.What are the magnitude and direction of the electric field at the point in question?
B.<span>What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field?"</span>
<span>Solution:
A) A charge q </span>under an electric field of intensity E will experience a force F equal to:

In our problem we have
and
, so we can find the magnitude of the electric field:

The charge is negative, therefore it moves against the direction of the field lines. If the force is pushing down the charge, then the electric field lines go upward.
B) The proton charge is equal to

Therefore, the magnitude of the force acting on the proton will be

And since the proton has positive charge, the verse of the force is the same as the verse of the field, so upward.
Here we will say that there is no external torque on the system so we will have

here we know that

where we know that

Also we know that

initial angular speed will be

now from above equation



now we have

so final speed will be 2.41 rad/s