qa = 1 nC, located 1 cm from B; qb = -1 nC, located 1 cm from C; qc = 4 nC. What is the magnitude of the electric force on charge A?

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Study for the UCF PHY2054 General Physics Exam. Use flashcards and multiple-choice questions complete with hints and explanations. Boost your understanding and get exam-ready!

Multiple Choice

qa = 1 nC, located 1 cm from B; qb = -1 nC, located 1 cm from C; qc = 4 nC. What is the magnitude of the electric force on charge A?

Explanation:
Electric forces add vectorially, so the force on A is the sum of the forces from B and from C. Compute each force: from B, F_AB = k|q_A q_B|/r_AB^2 = (9×10^9)(1×10^-9)(1×10^-9)/(0.01)^2 = 9×10^-5 N. Since A and B have opposite charges, this force pulls A toward B (to the right). From C, F_AC = k|q_A q_C|/r_AC^2. with r_AC = 0.02 m, this is (9×10^9)(1×10^-9)(4×10^-9)/(0.02)^2 = 9×10^-5 N. A and C have the same sign, so this force pushes A away from C (to the left). These two forces have equal magnitude and opposite directions, so they cancel. The net force on A is 0 N.

Electric forces add vectorially, so the force on A is the sum of the forces from B and from C.

Compute each force: from B, F_AB = k|q_A q_B|/r_AB^2 = (9×10^9)(1×10^-9)(1×10^-9)/(0.01)^2 = 9×10^-5 N. Since A and B have opposite charges, this force pulls A toward B (to the right).

From C, F_AC = k|q_A q_C|/r_AC^2. with r_AC = 0.02 m, this is (9×10^9)(1×10^-9)(4×10^-9)/(0.02)^2 = 9×10^-5 N. A and C have the same sign, so this force pushes A away from C (to the left).

These two forces have equal magnitude and opposite directions, so they cancel. The net force on A is 0 N.

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