Particle A of charge 3.06 10-4 C is at the origin, particle B of charge - 5.70 10-4 C is at (4.00 m, 0), and particle C of charge 1.08 10-4 C is at (0, 3.00 m). We wish to find the net electric force on C.

Question

Physics

Tianna Myers

25 November, 14:24

Particle A of charge 3.06 10-4 C is at the origin, particle B of charge - 5.70 10-4 C is at (4.00 m, 0), and particle C of charge 1.08 10-4 C is at (0, 3.00 m). We wish to find the net electric force on C.

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Answers (1)

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Angelica Goodman · Answer 1

F_net = 26.512 N

Explanation:

Given:

Q_a = 3.06 * 10^ (-4) C

Q_b = - 5.7 * 10^ (-4) C

Q_c = 1.08 * 10^ (-4) C

R_ac = 3 m

R_bc = sqrt (3^2 + 4^2) = 5m

k = 8.99 * 10^9

Coulomb's Law:

F_i = k * Q_i * Q_j / R_ij^2

Compute F_ac and F_bc:

F_ac = k * Q_a * Q_c / R^2_ac

F_ac = 8.99 * 10^9 * (3.06 * 10^ (-4)) * (1.08 * 10^ (-4)) / 3^2

F_ac = 33.01128 N

F_bc = k * Q_b * Q_c / R^2_bc

F_bc = 8.99 * 10^9 * (5.7 * 10^ (-4)) * (1.08 * 10^ (-4)) / 5^2

F_bc = - 22.137 N

Angle a is subtended between F_bc and y axis @ C

cos (a) = 3 / 5

sin (a) = 4 / 5

Compute F_net:

F_net = sqrt (F_x ^2 + F_y ^2)

F_x = sum of forces in x direction:

F_x = F_bc*sin (a) = 22.137 * (4/5) = 17.71 N

F_y = sum of forces in y direction:

F_y = - F_bc*cos (a) + F_ac = - 22.137 * (3/5) + 33.01128 = 19.72908 N

F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N

Answer: F_net = 26.512 N