Question

The -2.0 nC charge in the figure is in equilibrium. What is the charge of q ?

Answer 1

Hi there!

The easiest way to find the charge of q would be to find the "downward" (relative to the page) forces produced by the two positive 2.0nC charges. The force produced by 'q' would have to be equal and in the opposite direction to the sum of these forces.

The force equation involving two charges is:

`F_E = (kq_1q_2)/(r^2)`

`F_E` = Electrostatic Force (N)

k = Coulomb's Constant (8.99 × 10⁹ Nm²/C²)

q₁, q₂ = Charges (C)
r = distance between charges (m)

Let's begin by finding the electrostatic force between the -2.0 nC charge and one 2.0 nC charge.

(Recall that nC is 10⁻⁹ C and cm is 10⁻² m)

`F_E = ((8.99*10^(9))(2.0*10^(-9))(2.0*10^(-9)))/((√(0.03^2 + 0.02^2))^2)`

*We are using the Pythagorean theorem to find the total distance.

Plugging into a calculator:

`F_E = 2.767*10^(-5) N`

We are only interested in the VERTICAL component of this force, since the 'q' charge is directly in line with the -2.0nC charge and does not produce any horizontal forces.

First, let's find the angle that the 2.0nC charges make with the horizontal.

`tan^(-1)((2)/(3)) = 33.69^o`

Now, we can plug this value of the angle into sine to find the vertical component of the force.

`F_E(vertical) = F_E sin\theta = F_E sin(33.69) = 1.53 * 10^(-5) N`

Since there are two charges and the other has the SAME charge, we can simply double this quantity.

`F_E (\text{vertical, total}) = 3.07 *10^(-5) N`

So, for the -2.0 nC charge to be in equilibrium, the 'q' charge must produce an electrostatic force that is equal to the above total.

`F_E = (kq_1q)/(r^2)`

Rearrange to solve for q.

`q = (r^2 F_E)/(kq_1)`

Plug and solve.

`q = ((0.02^2) (3.07*10^(-5)))/((8.99*10^9)(2.0*10^(-9)))} = 6.83*10^(-10) C`

So:

`\large\boxed{q = 6.83*10^(-10) C = 0.68 nC}`