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 Q1 given the following: Q1 =-1 mC, Q2 =-5 mC, Q3 =-7 mc, r1 = 90 m, and r2 = 180 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 Q1? 4.69 in units of N) The total force on Q1 is going to be a vector sum of F12 (the force on Q1 due to Q2) and F13 (the force on Q1 due to Q3). You need to find both the magnitudes and directions of F12 and F13. The magnitudes of the forces are given by F12 kQ1Q/r12 and F13 kQ103/(ri+r2)2 where k-9e9 Vm/C and the absolute values for all charges should be used The direction of each force is determined by the two charges, where opposite charges attract and same charges repel Submit Answer Incorrect. Tries 1/3 Previous Tries What is the net force on charge Q2? (in units of N) Submit Answer Tries 0/3 What is the net force on charge Q3? in units of N) Submit Answer Tries 0/3 What is the sum of the forces on all three charges? 1.37 The total force on Q1 is going to be a vector sum of F12 (the force on Q1 due to Q2) and F13 (the force on Q1 due to Q3). You need to find both the magnitudes and directions of F12 and F13 The magnitudes of the forces are given by F12 kQ102/r12 and F13 kQ1Q3/(r1+r2), where k-9e9 Vm/C and the absolute values for all charges should be used The direction of each force is determined by the two charges, where opposite charges attract and same charges repel. Submit Answer Incorrect. Tries 1/3 Previous Tries (in units of N)

Explanation / Answer

Electrostatic force is given by:

F = k*Q1*Q2/R^2

Force is attractive if both charges have different signs and Force is repulsive if both forces have same sign.

Part A.

Net force on charge q1 will be

F12 is repulsive force towards left and F13 is repulsive force towards left, So

F1 = - F12 - F13

F1 = - k*q1*q2/r12^2 - k*q1*q3/r13^2

r12 = r1 = 90 m

r13 = r1 + r2 = 90 + 180 = 270 m

So,

F1 = [-9*10^9*1*10^-3*5*10^-3/90^2] - [9*10^9*1*10^-3*7*10^-3/270^2]

F1 = -6.42 N

Since F1 is negative, So direction of force is towards left.

Part B.

Net force on charge q2 will be

F23 is repulsive force towards left and F21 is repulsive force towards right, So

F2 = F21 - F23

F2 = k*q2*q1/r21^2 - k*q2*q3/r23^2

r21 = r1 = 90 m

r23 = r2 = 180 m

So,

F2 = [9*10^9*1*10^-3*5*10^-3/90^2] - [9*10^9*5*10^-3*7*10^-3/180^2]

F2 = -4.17 N

Since F2 is negative, So direction of force is towards left.

Part C.

Net force on charge q3 will be

F23 is repulsive force towards right and F31 is also repulsive force towards right, So

F3 = F31 + F23

F3 = k*q3*q1/r31^2 + k*q2*q3/r23^2

r31 = r1 + r2 = 270 m

r23 = r2 = 180 m

So,

F3 = [9*10^9*1*10^-3*7*10^-3/270^2] + [9*10^9*5*10^-3*7*10^-3/180^2]

F3 = 10.59 N

Since F3 is positive, So direction of force is towards right.

Part D.

Net force will be

Fnet = F1 + F2 + F3

Fnet = -6.42 - 4.17 + 10.59 = 0 N

Fnet = 0 N

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