Question

two charged particles are 1.2 m apart and experience a repulsive instantaneous electrostatic force of 4.3 x 10⁵ n. if one of the objects is charged at 3.1 mc, what is the charge (including the sign) of the other object?

Answer 1

Using Coulomb's Law, the charge of the second object was calculated by rearranging the formula and plugging in the given values of the force, the known charge, and the distance between the charges. We determine that the charge must be positive due to the repulsive force.

Step-by-step explanation:

The student's question revolves around determining the charge of a second object given the charge of the first object, the distance separating them, and the magnitude of the electrostatic force between them.

Coulomb's Law, which is used to calculate the electrostatic force (F) between two charges (q1 and q2) separated by a distance (r), is given by the equation:
F = k * |q1 * q2| / r²
where k is Coulomb's constant (8.988 x 10^9 Nm²/C²).

In this case, we know the following:
- F = 4.3 x 10^5 N
- q1 = 3.1 x 10^-6 C (since 1 microcoulomb (μC) equals 1 x 10^-6 C)
- r = 1.2 m

We can rearrange the equation to solve for q2:
q2 = F * r² / (k * q1)

Substituting the known values, we calculate for q2:

q2 = (4.3 x 10^5 N * (1.2 m)²) / (8.988 x 10^9 Nm²/C² * 3.1 x 10^-6 C)

Calculating this, we find the charge of the second object including its sign, which will be positive, as the force described is repulsive indicating that both charges must have like signs.