Answer:
A. -2.16 * 10^(-5) N
B. 9 * 10^(-7) N
Explanation:
Parameters given:
Distance between their centres, r = 0.3 m
Charge in first sphere, Q1 = 12 * 10^(-9) C
Charge in second sphere, Q2 = -18 * 10^(-9) C
A. Electrostatic force exerted on one sphere by the other is:
F = (k * Q1 * Q2) / r²
F = (9 * 10^9 * 12 * 10^(-9) * -18 * 10^(-9)) / 0.3²
F = -2.16 * 10^(-5) N
B. When they are brought in contact by a wire and are then in equilibrium, it means they have the same final charge. That means if we add the charges of both spheres and divided by two, we'll have the final charge of each sphere:
Q1 + Q2 = 12 * 10^(-9) + (-18 * 10^(-9))
= - 6 * 10^(-9) C
Dividing by two, we have that each sphere has a charge of -3 * 10^(-9) C
Hence the electrostatic force between them is:
F = [9 * 10^9 * (-3 * 10^(-9)) * (-3 * 10^(-9)] / 0.3²
F = 9 * 10^(-7) N
Acceleration is the increase in the rate of speed so if its acceleration is decreasing then I guess use negative.
hope this helps
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