Three point charges are arranged in a horizontal line as shown below. Find the e

Question

Three point charges are arranged in a horizontal line as shown below. Find the electric force on Q_1 given the following: Q_1 = -4 mC, Q_2 = 1 mC, Q_3 = -5 mC, r_i = 40 m, and r_2 = 60 m. Remember that a positive force points to the right and a negative force points to the left. What is the net force on charge Q_1? The total force on Qi is going to be a vector sum of F12 (the force on Qi due to Q_2) and F_13 (the force on Q_1 due to Q3). You need to find both the magnitudes and directions of F_12 and F_13. The magnitudes of the forces are given by:

Explanation / Answer

here,

Q1 = - 0.004 C

Q2 = 0.001 C

Q3 = - 0.005 C

r1 = 40 m

r2 = 60 m

the net force on charge Q1 , F1 = k * Q1*Q2/r1^2 - k*Q1*Q3/(r1 + r2)^2

F1 = 9 * 10^9 * (0.004 * 0.001/40^2 - 0.004 * 0.005 /( 40 +60)^2 )

F1 = 4.5 to the right

the net force on charge Q2 , F2 = - k * Q1*Q2/r1^2 + k*Q2*Q3/(r2)^2

F2 = 9 * 10^9 * (-0.004 * 0.001/40^2 + 0.001 * 0.005 /( 60)^2 )

F2 = - 10 N

the force F2 is 10 N to the left

the net force on charge Q3 , F3 = k * Q1*Q3/(r1 + r2)^2 - k*Q2*Q3/(r2)^2

F3 = 9 * 10^9 * (0.004 * 0.005 /(40 +60)^2 + 0.001 * 0.005 /( 60)^2 )

F3 = 30.5 N

the force F3 is 30.50 N to the right

the sum of all force , F = F1 + F2 + F3

F = 4.5 - 10 + 30.5

F = 25 N

the sum of all forces is 25 N

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