# Two fixed charges of 1.0uC and 3.0uC are 15cm apart. Where would you place a third charge...

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Two fixed charges of 1.0uC and 3.0uC are 15cm apart. Where would you place a
third charge so that no net electrostatic force acts on it?

The Us before the Cs represent micro. I know how to use coulombs law for 2 charges but here I am lost, please help.

asked Mar 5, 2014

We want to find the place were the net force on the third charge is zero. Essentially we are using Coulombs law twice (once with charges 1 and 3, and then once with charges 2 and 3) and ensuring that it is equal both times by carefully placing the third particleTo do so we may set up Newton's equation (The particles should be in a line so only one-dimension):

$$\Sigma F_3 = 0 = F_1 + F_2 = \frac{ k q_1 q_3}{ r_1^2} - \frac{k q_2 q_3}{r_2^2}$$

Here $r_1$ is the distance between charge 1 and charge 3, and $r_2$ is the distance between charge 2 and charge 3. We know that $r_1 + r_2 = d$ according to the problem statement ( $d= 15 cm$ ). The minus sign in front of the force from charge 2 signifies that it is acting in the opposite direction of charge 1.

Dividing out the common factors and a bit of algebra:

$$\frac{q_1}{r_1^2} = \frac{q_2}{(d - r_1)^2}$$

$$q_1 (d^2 -2 d r_1 + r_1^2) = q_2 r_1^2$$

$$(q_1 - q_2) r_1^2 - 2 q_1 d r_1 + q_1 d^2 = 0$$

Here we are left with a quadratic equation which can be solved using the quadratic formula leaving us with:

$$r_1 = 5.5 cm$$

Note that the quadratic equation will yield two answers...you need to choose which makes sense in this situation. In this case, only one is between 0 and 15 cm and that is your answer.
answered Mar 13, 2014 by (2,140 points)