A point charge q 1 = -2.3 C is located at the origin of a co-ordinate system. An

Question

A point charge q1 = -2.3 C is located at the origin of a co-ordinate system. Another point charge q2 = 6.4 C is located along the x-axis at a distance x2 = 6.2 cm from q1.

2) Charge q2 is now displaced a distance y2 = 2.4 cm in the positive y-direction. What is the new value for the x-component of the force that q1 exerts on q2?

3) A third point charge q3 is now positioned halfway between q1 and q2. The net force on q2 now has a magnitude of F2,net= 9.991 N and points away from q1 and q3. What is the value (sign and magnitude) of the charge q3?

Explanation / Answer

For charges Q1 and Q2 separated by a distance r, the net force acting between them is given by kQ1*Q2 / r^2

2.) As the charge, after being displaced, is now located at (6.2 , 2.4) and q1 is at origin, the distance between the charges is given as sqrt(6.2^2 + 2.4^2)

Also, Q1 = -2.3 uC and Q2 = 6.4 uC that is the charges will pull each other, hence the x component of the force that Q1 exerts on Q2 will be along - x axis.

That is, F = kQ1 * Q2 / r^2 = - 8.99 x 10^9 x 2.3 x 6.4 x 10^-12 / (6.2^2 + 2.4^2)*10^-4 = -29.939 N

Fx = - 8.99 x 10^9 x 2.3 x 6.4 x 10^-12 x Cosa / (6.2^2 + 2.4^2)*10^-4 where cos a = 6.2 / sqrt(6.2^2 + 2.4^2)

Hence Fx = - 132.3328 x 10^-3 x 6.2 / (6.2^2 + 2.4^2)x10^-4*sqrt(6.2^2 + 2.4^2)

or Fx = - 820.46336 x 10 / 44.2 * 6.648 = 27.922 N

3.) Now the net force acting on Q2 after placing Q3 in between is 9.991 N and points away from Q1 and Q2.

We know that Q1 is pulling Q2 by a force of 29.939, so the Q3 must repel it with a net force of 39.93 N

That is KQ3*Q2 / r^2 = 39.93

or, 8.99 x 10^3 x 6.4 x Q3 / 10^-4*(44.2/4) = 39.93

5.2069 x 10^7 x Q3 = 39.93

or, Q3 = 7.669 x 10^-7 C

Q3 must be a positive charge so as to repel Q2 which itself is positive.

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