finding magnitude of charged particle

Question

positively charged particle Q1 = +45 nC is held fixed at the origin. A second charge Q2 of mass m = 7.5 μg is floating a distance d = 25 cm above charge Q1. The net force on Q2 is equal to zero. You may assume this system is close to the surface of the Earth.

Calculate the magnitude of Q2 in units of nanocoulombs.

Answer 1

Hello Julie!

A: The NET force on q2 is zero; therefore, the repulsive coulombic force ( Fc ) acting upward on q2 must be balanced out by the force of gravity ( Fg ) exerted downward by the earth’s gravitational pull. If ( Fc – Fg ) = 0, then ( Fc ) = ( Fg ).

Note: The charge on q1 has to be positive in order for the force between the two particles to be repulsive.

The force of gravitational attraction upon an object very close to the earth’s surface is ( Fg ) = ( GM1m2 / r2 ), where ( G ) = the universal gravitational constant of attraction, ( M1 ) is the mass of the earth, ( m2 ) = the mass of the suspended particle, and ( r ) = the radius of the earth ( distance from the earth’s center to the earth’s surface ). If we assume that q1 rests upon the earth’s surface, we can assume that its mass is negligible, and the gravitational force exerted upon q2 is primarily due to the mass of the earth. The aforementioned equation can be rearranged as follows:

Fg = ( m2 )(GM1/ r2 ). Close to the earth’s surface, the values of G, M1, and r are constant, and (GM1/ r2 ) = 9.8 m/s2. Therefore, the downward force acting upon q2 is Fg = ( m2 )( g ). Setting Fc = Fg, the net force exerted upon q2 is ( k )( q1q2 / r2 ) = ( m )( g ), where ( r ) = 25 cm ( 2.5*10-1 m ).

q2 = [ ( m2g )( r2 ) ] / [ ( k )( q1 ) ].

q2 = [ ( 7.5*10-9 kg )( 9.8 m / s2 )( 2.5*10-1 m )2 ] / [ ( 8.9*109 N*m2/C2 )( + 4.5*10-8 C ) ].

q2 = ( 5.0*10-9 C2 ) / ( 15.8 C ) ) = 3.16*10-10 C = 0.316*10-9 C

q2 = 0.316 nC.