# 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.

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)