Find the electrical charges of q1 and q2 with force of 10 N.

Question

Two identical small metal spheres 3 cm apart attract each other with an electrical force of 150 N. They are temporarily connected by a conducting wire, which is then removed.

Determine the initial electrical charges if they now repel each other with a force of 10 N. (It is assumed that the charge on each sphere is uniformly distributed).

(Answer: q1 = ± 5.00 μC; q2 = ± 3.00 μC)

Answer 1

In the initial state, we have:

F = kq₁q₂ / r² = -150N

We use a negative force (F = -150N) to indicate that the force is attractive. We know k (9*10⁹ Nm²/C²) and r (3cm), so we solve for q₁q₂:

q₁q₂ = Fr²/k = (-150N)(0.03m)²/(9*10⁹ Nm²/C²) = -15µC

In the final state, since they were touched by a conducting wire, the two spheres must have equal charge (call it capital Q).

F` = kQ² / r² = 10N

Again, we can solve for Q:

Q²= F`r²/k = (10N)(0.03m)²/(9*10⁹ Nm²/C²)

Q = ±1µC

The temporary connection of the metal wire is an event that must conserve the charge of the spheres. Hence: q₁+q₂ = Q+Q = ±2µC

So, we have two equations and two unknowns:

q₁ + q₂ = ±2

q₁q₂ = -15

Solving thi system yields the solutions:

q₁ = 5µC q₂ = -3µC

q₁ = -5µC q₂ = 3µC

q₁ = 3µC q₂ = -5µC

q₁ = -3µC q₂ = 5µC