Answer:

Explanation:
Given:
- quantity of point charge,

- radial distance from the linear charge,

- linear charge density,

<u>We know that the electric field by the linear charge is given as:</u>



<u>Now the force on the given charge can be given as:</u>



Answer:
1.5F
Explanation:
Using
E= F/q
Where F= force
E= electric field
q=charge
F= Eq
So if qis tripled and E is halved we have
F= (E/2)3q
F= 1.5Eq=>> 1.5F
I think its suicidal ideation......
I think
The magnetizing current in a transformer is rich in 3rd harmonic. This is because harmonics are AC voltages and currents with frequencies that are generally higher.
Nothing happens to the mirror.
However, if the ray is within some suitable range of
wavelengths, the ray is reflected from the mirror's surface.