A -4.4 nC charge exerts a repulsive force of 36 mN on a second charge which is located a distance of 0.70 m away from it.
What is the charge (magnitude and sign) of the second charge?
To find the charge (magnitude and sign) of the second charge, we can use Coulomb's law.
Coulomb's law states that the magnitude of the electrostatic force between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's law is:
F = k * (|q1| * |q2|) / r^2
Where:
F is the electrostatic force between the two charges,
k is Coulomb's constant (k = 8.99 x 10^9 N m^2/C^2),
|q1| and |q2| are the magnitudes of the charges,
and r is the distance between the charges.
In this case, we are given:
F = 36 mN = 36 x 10^-3 N (since 1 mN = 10^-3 N),
|q1| = 4.4 nC = 4.4 x 10^-9 C (since 1 nC = 10^-9 C),
and r = 0.70 m.
Using Coulomb's law, we can rearrange the formula to solve for |q2|:
|q2| = (F * r^2) / (k * |q1|)
Substituting the given values into the formula, we get:
|q2| = (36 x 10^-3 N * (0.70 m)^2) / (8.99 x 10^9 N m^2/C^2 * 4.4 x 10^-9 C)
Simplifying the expression, we find:
|q2| = 1.12 x 10^-5 C
Therefore, the magnitude of the second charge is 1.12 x 10^-5 C.
Since the first charge is negative (-4.4 nC), and it exerts a repulsive force on the second charge, we can infer that the second charge must be negative as well.
Hence, the charge (magnitude and sign) of the second charge is -1.12 x 10^-5 C.