Question

Consider three point charges at the corners of an equilateral triangle, its base sitting on the x-axis, with charge qA positioned at the origin, qB is at x=+3.50 cm, and qC is at the (top) third corner. qC = +2.60 μC, and the net electric force on qC is 375.0 N in the negative y direction. a. Explain briefly but clearly how you know that qA and qB must have the same magnitude charge and what the sign of that charge must be. b. Find the sign and magnitude of the charges on qA and qB.

Answer 1

part a )

By applying coulomb law for electrostatic force , the net force is point downwards , it means both charges are negative sign and have the same magnitude . The diagram shows the attractive force F from each negative charge directed along the lines between the charges. Only when each force has the same magnitude (which is the case when both unknown charges have the same magnitude) will the resultant force point vertically downward. This occurs because the horizontal components of the forces cancel, one pointing to the right and the other to the left (see the diagram). The vertical components reinforce to give the observed downward net force.

part b )

by part a we know that unknown charges have same magnitude

F = kQa*(2.6 x 10^-6 C) / r^2 = KQb(2.6x10^-6)/r^2

The magnitude of the net force acting on the 2.6 micro C charge, then, is the sum of the magnitudes of the two vertical components F cos 30degree

sum of F = K|Qa|(2.6 x 10^-6C)*cos30/r^2 + K|Qb|(2.6 x 10^-6 C)*cos30/r^2

since Qa = Qb

sum of F = 2k|Qa|*(2.6 x 10^-6 )cos30 / r^2

here

sum of F = 375 N

r = 3.5 x 10^-2 m

k = 9 x 10^9

Qa = (sum of F) * r^2 / ( 2k * cos30 * 2.60 x 10^-6)

Qa = (375 ) ( 3.5 x 10^-2)^2 / ( 2 x 9 x 10^9 x cos30 x 2.6 x 10^-6)

Qa = 11.3 x 10^-6 C

and Qa = Qb = -11.3 x 10^-6 C