Electric forces

Question

The figure below shows two identical small, charged spheres. One of mass 3.6 g is hanging by an insulating thread of length 20.0 cm. The other is held in place and has charge q₁ = −4.9 µC.

 The thread makes an angle of 18° with the vertical, resulting in the spheres being aligned horizontally, a distance r apart. Determine the charge q₂ on the hanging sphere.

 nC

Answer 1

First, let’s determine the value of r. According to trigonometry, r = 20 cm x sin 18 = 6.18 cm or 0.0618m.

Now let’s determine the forces. There is a force to the right of q2 that is equal to the coulomb force between q1 andd q2 or Fc = K q1 x q2 / r² where K = 8.99 x 10⁹ Nm²/C².

Fc = 8.99 x 10⁹ x -4.9 x 10^-6C x -q2 / (6.18 x 10^-2m)² = 1.15 x 10⁷ x q2 N. Since q2 is stationary, there must be a force on q2 acting to the left. This is equal to -Tcos(90-18)° where T = the tension in the string. The sum of the forces in the X direction must equal zero, so

1.15 x 10⁷q2 -Tcos72 = 0. As there are 2 unknowns, we need another equation.

The downward force of gravity on the charge is -3.6g x 9.8 m/s² that must be balanced by an upward force given by Tsin72 or Tsin72 – 3.6 x 10^-3 m 9.8 m/s² = 0. Now we can solve this for T and get T = 3.71 x 10^-2 N.

Plug T into the equation in bold above and solve for q2. Q2 = 1.0 x 10^-9C or 1.0 nC

Check all calculations as it is easy to make math errors when there are multiple calculations!