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Three point charges are arranged on a line. Charge q3=+5.00nCand is at the origin. Charge q2=-3.00nCand is at x=+4.00cm. Charge q1is atx=+2.00cm . What is q1 (magnitude and sign) if the net force on q3is zero?

Short Answer

Expert verified

The magnitude and sign of q1 is +0.75nC.

Step by step solution

01

Coulomb’s law

Coulomb's law states that the magnitude F of the force exerted by two-point charges q1 and q2 separated by is directly proportional to the product of the charges(q1×q2) and inversely proportional to the square of the distance between them.

F=kq1q2r2

Here; F is the force on each point charge exerting on each other, k is the proportionality constant, r is the distance between the charges q1 and q2.

02

Magnitude and sign of q1 is

Consider the given data as below.

The charge,q2=-3.00nC=-3.00×10-9C

The charge, q3=+5.00nC=+5.00×10-9C

The force, F = 0.600 N

The distance between two point charges,

The Coulomb’s constant,k=9×109N·m2/C2

Draw the free body diagram as below.

The charge exerts a force on charge q3, the net force acting on q3 is zero.

As q3 is affected by two forces F13 and F23, thus, the net force is;

Fnet=F13+F23

Thus, the charge q1 towards the positive x-direction, which is a repulsive force between q3 and q1, so the net force is;

F13=F23kq1q3r12=kq2q3r22q1r12=q2r22q1=q2r12r22

Substitute known values ion the above equation, and you have

q1=∣3.00nCl2.00cm4.00cm2=+3.00nC4.00=+0.75nC

Hence, the magnitude and sign of q1 is +0.75nC.

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