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 = 3 mC, Q2 = 5 mC, Q3 =-3 mC, 70 m, and r2 = 105 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? (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 kQ1Q2/r12 and F13 = kQ1Q3/(r1+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 1/2 Previous Tries What is the net force on charge Q2? (in units of N) Submit AnswerTries 0/2 What is the net force on charge Q3?

Explanation / Answer

Force on q1 by q2 is F12 = k*q1*q2/r1^2 = (9*10^9*3*10^-3*5*10^-3)/70^2 = 27.55 N = -27.55 N

Force on q1 by q3 is F13 = k*q1*q3/(r1+r2)^2 = (9*10^9*3*10^-3*3*10^-3)/(105+70)^2 = 2.65 N

so the net force on Q1 is -27.55+2.65 = -24.9 N

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force on q2 by q1 is F21 = - F12= 27.55 N

Force on q2 by q3 is F23 = (k*q2*q3/r2^2) = (9*10^9*5*10^-3*3*10^-3)/105^2 = 12.25 N

Net force on Q2 is 27.55+12.25 = 39.8 N

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Force on q3 by q1 is F31 = -F13 = -2.65 N

Force on q3 by q2 is F32 = -F23 = -12.25

Net Force on q3 is -2.65-12.25 = -14.9 N

sum of the forces on all three charges is -24.9+39.8-14.9 = 0 N

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