A negative charge of - 0.0005 C exerts an attractive force of 19.0 N on a second charge that is 25 m away. What is the magnitude of the second charge?

0.002638 C or 2.6388*10^-3 C

Explanation:

Given that

Quantity of the charge, q = -0.0005 C

Force on the charge of magnitude, F1 = 19 N

Distance from the second charge, r = 25 m

Magnitude of force of the second charge, q2 = ? N

F = (kq1q2) / r², where

k = 9*10^9

19 = (9*10^9 * 0.0005 * q2) / 25²

19 * 625 = 4.5*10^6 * q2

q2 = 11875 / 4.5*10^6

q2 = 0.002638 C or 2.6388*10^-3 C

Thus, the magnitude of the second charge is 0.002638 C or 2.6388*10^-3 C

pe = weightxheight = 2000x1000=2,000,000joules

i think its d if wrong im sorry