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How does the electric potential energy between two charged particles change if one particle's charge is increased by a factor of 2?

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

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