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 forces (in units of kQ2/R2) on the charges given the following: Q1 = 48 Q, Q2-4 Q, Q3 -180 Q, r1-4 R, and r2 2 R. Remember that a positive force points to the right and a negative force points to the left. Q3 What is the net force on charge Q1? kQ2/R2 252 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 kQ103/(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 2/3 Previous Tries What is the net force on charge Q2? 168 The total force on Q2 is going to be a vector sum of F21 (the force on Q2 due to Q1) and F23 (the force on Q2 due to Q3). You need to find both the magnitudes and directions of F21 and F23 The magnitudes of the forces are given by: F21 kQ2Q1/r1 and F23 kQ23/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 AnswerIncorrect. Tries 2/3 Previous Tries What is the net force on charge Q3? kQ2/R2 Submit Answer Tries 0/3

Explanation / Answer

Q1 = 48 Q

Q2 = 4Q

Q3 = -180 Q

r1 = 4R

r2 = 2R

force on Q1

force on Q1 due to Q2 , F21x = - k*Q2*Q1/r1^2 = -4*48/4^2 kQ^2/R^2 = -12 k*Q^2/R^2

force on Q1 due to Q3 , F31x = + k*Q3*Q1/(r1+r2)^2 = 180*48/6^2 kQ^2/R^2 = 240 k*Q^2/R^2

net force on Q1 = F21 + F31x = 228 kQ^2/R^2

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force on Q2

force on Q2 due to Q1 , F12x = k*Q1*Q2/r1^2 = 48*4/4^2 kQ^2/R^2 = 12 k*Q^2/R^2

force on Q2 due to Q3 , F32x = + k*Q3*Q2/r2^2 = 180*4/2^2 kQ^2/R^2 = 180 k*Q^2/R^2

net force on Q3 = F12x + F32x = 192 kQ^2/R^2

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force on Q3

force on Q3 due to Q1 , F13x = - k*Q1*Q3/(r1+r2)^2 = -48*180/6^2 kQ^2/R^2 = -240 k*Q^2/R^2

force on Q3 due to Q2 , F23x = - k*Q2*Q2/r2^2 = -180*4/2^2 kQ^2/R^2 = -180 k*Q^2/R^2

net force on Q3 = F13x + F23x = -420 kQ^2/R^2

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