The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.
The electric potential energy of the charge is equal to the potential at the location of the charge, V, times the charge, q:

The potential is given by the magnitude of the electric field, E, times the distance, d:

So we have

(1)
However, the electric field is equal to the electrical force F divided by the charge q:

Therefore (1) becomes

And if we use the data of the problem, we can calculate the electrical potential energy of the charge:
Because the temperature causes it to go the physical change like if water is froze the temperate just caused it to go through another state of matter another example as if the ice got melted you would need high temperature to cause this change physically. Let me know if I help or if their is anything I can change for you to understand better
Answer: q = -52.5 μC
Explanation:
The complete question is given thus;
A point charge Q moves on the x-axis in the positive direction with a speed of 280 m/s. A point P is on the y-axis at y=+70mm. The magnetic field produced at the point P, as the charge moves through the origin, is equal to -0.30uTk. What is the charge Q? (uo=4pi x 10^-7 T m/A).
SOLVING:
from the given parameters we can solve this problem.
Given that the
Speed = 280 m/s
y = 70mm
B = -30 * 10⁻⁶T
Using the equation for magnetic field we have;
Β = μqv*r / 4πr²
making q (charge) the subject of formula we have that;
q = B * 4 *πr² / μqv*r
substituting the values gives us
q = (-0.3*10⁻⁶Tk * 4π * 0.07²) / (4π*10⁻⁷ * 280 ) = - [14.7 * 10⁻¹⁰k / 2.8 * 10⁻⁵ k ]
q = -52.5 μC
cheers i hope this helped !!!