Question

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?

Answer 1

Answer:

0.00264C

Explanation:

Using the Columbs law formula

F = kq1q2/r²

F is the force of attraction

q1 and q2 are the charges

r is the distance between the charges.

k is the Coulomb's cinsrsnty

Given F = 19N, q1 = -0.0005C

r = 25m q2 = ?

Substituting

19 = 9×10^9×-0.0005q2/25²

11875 = -4,500,000q2

q2 = 11,875/-4,500,000

q2 = - 0.0264C

Magnitude of the second charge is 0.00264C

Answer 2

Answer:

0.00264 C

Explanation:

From Coulomb's law,

F = kqq'/r²........................ Equation 1

Where F = Force of attraction, k = coulombs constant, q = second charge, q' = First charge.

make q the subject of the equation

q = Fr²/kq'.................... Equation 2

Given: F = 19 N, r = 25 m, q = 0.0005 C, k = 9×10⁹ Nm²/C²

q = 19×25²/(0.0005×9×10⁹)

q = 11875/4500000

q = 0.00264 C

Hence the magnitude of the second charge = 0.00264 C