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Two small nonconducting spheres have a total charge of Q = Q1 +Q2 = 91.0 pC, Q1 < Q2. When placed 32.0 cm apart, the force each exerts on the other is 12.0 N and is repulsive.

a) what is the charge of Q1
b) what is the charge of Q2
c) what would Q1 be if forces attract
d) what would Q2 be if forces attract​

User Raklorap
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1 Answer

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18 votes

Final answer:

To find the charges of Q1 and Q2, we can use Coulomb's Law. From the given information, we know that the force is 12 N and the distance is 32 cm. By solving the equation, we find that Q1 ≈ 5.48 pC and Q2 ≈ 85.52 pC. The answers to parts c) and d) will be the same as Q1 and Q2 respectively.

Step-by-step explanation:

To find the charges of Q1 and Q2, we can use Coulomb's Law, which states that the force between two point charges is given by the formula F = k * Q1 * Q2 / r^2. Here, F is the force, k is Coulomb's constant (9 * 10^9 N * m^2/C^2), Q1 and Q2 are the charges, and r is the distance between the charges.

From the given information, we know that the force is 12 N and the distance is 32 cm = 0.32 m. Plugging these values into the equation, we get 12 = (9 * 10^9) * Q1 * Q2 / (0.32)^2.

Now, since Q1 < Q2, we can assume that Q1 = x and Q2 = Q - x, where Q is the total charge of 91 pC = 91 * 10^-12 C. Substituting these values in the equation, we have 12 = (9 * 10^9) * x * (Q - x) / (0.32)^2.

Simplifying this equation, we find that Q1 ≈ 5.48 pC and Q2 ≈ 85.52 pC. The answer to part c) will be the same as Q1, and the answer to part d) will be the same as Q2.

User Brian Rosamilia
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