Three charges, Q 1 , Q 2 , and Q 3 are located in a straight line. The position

Question

Three charges, Q1, Q2, and Q3 are located in a straight line. The position of Q2 is 0.312 m to the right of Q1. Q3 is located 0.158 m to the right of Q2.

In the above problem, Q1= 1.39·10-6 C, Q2= -2.65·10-6 C, and Q3= 3.18·10-6 C. Calculate the total force on Q2. Give with the plus sign for a force directed to the right.

Now the charges Q1= 1.39·10-6 C and Q2= -2.65·10-6 C are fixed at their positions, distance 0.312 m apart, and the charge Q3= 3.18·10-6 C is moved along the straight line. For what position of Q3 relative to Q1 is the net force on Q3 due to Q1 and Q2 zero? Use the plus sign for Q3 to the right of Q1.

Explanation / Answer

Here ,

as the electric force between two charges is given as

F = k * Q1*Q2/d^2

for the net force on the charge Q2 ,

Fnet = F3 - F1

Fnet = 9*10^9 * 2.65 *10^(-6) * 10^-6 *(3.18/0.158^2 - 1.39/.312^2)

calculating

Fnet = 2.698 N

the net force acting on the charge Q2 is 2.698 N

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let the position of the charge Q3 is x = -a from the charge Q1 ,

Now, as the force on the charge is zero where net electric field due to Q1 and Q2 is zero

E1 = E2

k * 1.39 *10^-6 /(a^2) = k * 2.65 * 10^-6 /( a + 0.312)^2

1.39 *10^-6 /(a^2) = 2.65 * 10^-6 /( a + 0.312)^2

solving for a

a = 0.819 m

position of Q3 = -a

position of Q3 = -0.819 m

Charge Q3 must be placed at -0.819 m relative to Q1

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