Three charges, Q_1, Q_2, and Q_3 are located in a straight line. The position of

Question

Three charges, Q_1, Q_2, and Q_3 are located in a straight line. The position of Q_2 is 0.257 m to the right of Q_1. Q_3 is located 0.136 m to the right of Q_2. The force on Q_2 due to its interaction with Q_3 is directed to the..... In the above problem, Q_1 = 1.56-10^-6 C, Q_2 = -2.84 10^-6 C, and Q_3 = 3.33-10^-6 C. Calculate the total force on Q_2. Give with the plus sign for a force directed to the right. Now the charges 1.56.10^-6 C and Q_2 = -2.84.10^-6 C are fixed at their positions, distance 0.257 m apart, and the charge Q_3 = 3.33.10^-6 C is moved along the straight line. For what position of Q_3 relative to Q_1 is the net force on Q_3 due to Q_1 and Q_2 zero? Use the plus sign for Q_3 to the right of Q_1.

Explanation / Answer

Given

   the position of charges

   Q1       Q2   Q3
   <-----0.257m----><-0.136m->

the force on Q2 due to interaction with Q3 is dircted towards left if two charges are positive

and
  

the force on Q2 due to interaction with Q3 is dircted towards left if two charges are negative

and it would be to the right if Q2 +ve charge and Q3 negative charge

2)

Q1= 1.56*10^-6 C

Q2= -2.84*10^-6 C

Q3= 3.33*10^-6 C

now force on Q2 is

   F12 + F32
F12 = kQ1*Q2/r1^2

= 9*10^9*1.56*10^-6 *2.84*10^-6 /(0.257)^2 N
= 0.6037 N
= -0.6037 N

F32 = kQ1*Q2/r1^2

= 9*10^9*3.33*10^-6 *2.84*10^-6 /(0.257)^2 N

= 1.28866 N

net force on Q2 is F = -0.6037 +1.28866 N = 0.68496 N +ve x direction
3)

   let Q3 is at beyond Q2 , x m from Q1 then , taking origin as at Q1

then

   F13 = F23

F13 = kQ1Q3/r13^2 ,               

= 9*10^9*1.56*10^-6*3.33*10^-6/(0.257+x)^2,
F23 = kQ2*Q3/r23^2

= -9*10^9*2.84*10^-6*3.33*10^-6/(x)^2

equating these then

   9*10^9*1.56*10^-6*3.33*10^-6/(0.257+x)^2 = -9*10^9*2.84*10^-6*3.33*10^-6/(x)^2

9*10^9*1.56*10^-6*3.33*10^-6/(0.257+x)^2 /9*10^9*2.84*10^-6*3.33*10^-6/(x)^2=1

solving for x = 0.02487 m

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