Sagot :
from the formula of electric potential = (1/4πe)*(Qq/r), if one of the charge is doubled, the electric potential energy would be doubled too. Not so sure though, u might wanna double-check with someone else. But hope that helps. :)
Answer:
Magnitude of potential energy is increased by factor "2"
Explanation:
As we know that if two charge particles are placed at some distance "r" from each other then the electrostatic potential energy between two charge particles is given as
[tex]U = \frac{kq_1q_2}{r}[/tex]
now we know that if the charge of one of the charge particle is increased to twice of initial charge then
[tex]U' = \frac{kq_1(2q_2)}{r}[/tex]
now we can say from above two equations that
[tex]U' = 2U[/tex]
so on increase one of the charge to twice of initial value then the potential energy will become TWICE