Two charged particles, Q_1 and Q_2 each carrying the elementary charge of 1.6 ti

Question

Two charged particles, Q_1 and Q_2 each carrying the elementary charge of 1.6 times 10^-19 C. are a distance of 15.30 times 10^-12 m apart. Find the force that Q_1 exerts on Q_2. Find the force that Q_2 exerts on Q_1. If the distance between the charges is increased to 1.7 times their initial separation, how is the force between them changed? Conceptualize Use the Active Figure to visualize the situation. The two charges are a specified distance apart, exerting forces on each other that are proportional to the product of their charges and inversely proportional to the square of the distance between them. The direction of the force is along the line joining the particles. Categorize This is a problem in applying Coulomb's law to calculate the electric force between two charges. Analyze (A) Find the force that Q_1 exerts on Q_2. The force of the first charge on the second charge, displaced by r from it, is F_12 = k_e q_1 q_2/r^2 f. Assume the vector r is along the x-axis. The force on the second particle points away from the first particle, and is repulsive. Evaluate the magnitude of the force as F_12 = (8.99 times 10^-9 N middot m^2/C^2) (1.6 times 10^-9 C) (1.6 times 10^-9 C)/(15.30 times 10^-12 m)^2 Find the force that Q_2 exerts on Q_1. The displacement vector from Q_2 to Q_1 is in the -r direction. Coulomb's law predicts a force of Q_2 exerted by Q_1 having the same magnitude as in part (A) for the force of Q_1 on Q_2, but in the -r direction, consistent with Newton's third law. If the distance between the charges is increased to 1.7 times their initial separation, how is the force between them changed? Replace r by (1.7)r in Coulomb's law to obtain F_12, new = k_e new = k_e q_1q_2/(1, 7)^2 r^2 = F_12/(1.7) = Finalize Notice the magnitude of the force that results from charges of 10^-19 C. Does this suggest that one coulomb is a very small or very large unbalanced charge?

Explanation / Answer

1) f12 = kq1q2/r^2 = 9*10^9*1.6*10^-19*1.6*10^-19/15.3^2*10^-24 = 0.09907*10^-5 N

2) according to newton's third law f12 = f21 ( only magnitudes but not the directions)

3) f12 new = f12/(1.7^2) = 0.9907*10^-5/1.7^2 = 0.343*10^-5 N

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