Question

A negative charge of - 8.0 x 10^-6 C exerts an attractive force of 12 N on a second charge that is

0.050 m away. What is the magnitude of the second charge?

Answer 1

Answer:

Magnitude of the second charge is $-4.17*10^{-7}C$

Explanation:

According to columbs law;

F = $kq1q2/r^{2}$

F is the attractive or repulsive force between the charges = 12N

q1 and q2 are the charges

let q1 = - 8.0 x 10^-6 C

q2=?

r is the distance between the charges = 0.050m

k is the coulumbs constant =9*10⁹ kg⋅m³⋅s⁻⁴⋅A⁻²

On substituting the given values

12 = 9*10⁹*( - 8.0 x 10^-6)q2/0.050²

Cross multiplying

$0.03=9*10^{9}* -8.0*10^{-6} q2\\0.03 = -72*10^{3} q2\\q2 = \frac{0.03}{ -72*10^{3}} \\q2 = -4.17*10^{-7}C$

Answer 2

Answer:

q₂ = + 4.17 x 10⁺⁷ C

Explanation:

From Coulomb's Law, we know that force of attraction or repulsion between two charges is given by:

F = kq₁q₂/r²

where,

F = force = 12 N

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of 1st charge = 8 x 10⁻⁶ C

r = distance between charges = 0.05 m

q₂ = magnitude of second charge = ?

Therefore,

12 N = (9 x 10⁹ Nm²/C²)(8 x 10⁻⁶ C)(q₂)/(0.05 m)²

0.03 Nm² = (72000 Nm²/C²)(q₂)

q₂ = (0.03 Nm²)/(72000 Nm²/C²)

q₂ = + 4.17 x 10⁺⁷ C

It is a positive charge, because there is attraction between opposite charges only.