Question

Learning goal: to practice problem-solving strategy 20.1 electric forces and coulomb's law. two charged particles, with charges q1=q and q2=4q, are located at a distance d=2cm apart on the x axis. a third charged particle, with charge q3=q, is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3. find the position of charge 3 when q = 1.0 nc . note that you are given the magnitude of q, but the sign of the charge q is not specified. q can be positive or negative, so that the three charges (q1, q2, q3) are either all positive charges or all negative charges.

Answer 1

Refer to the figure shown below.

d =  the distance between q₁ and q₂.
Let r =  the distance between q₂  and q₃.

Given:
d = 2 cm = 0.02 m
q₁ = q
q₂ = 4q
q₃ = q = 1.0 nC.

The forces between the charges are
$F_{13} = \frac{q_{1}q_{3}}{4 \pi \epsilon_{0}(d+r)^{2}} = \frac{q^{2}}{4 \pi \epsilon_{0}(d+r)^{2}}$
$F_{23} = \frac{(4q)q}{4 \pi \epsilon_{0}r^{2}}$

Because F₁₃ = F₂₃, therefore
$\frac{1}{(d+r)^{2}} = \frac{4}{r^{2}}\\ 4(d^{2}+2dr+r^{2})=r^{2} \\3r^{2}+8dr+4d^{2}=0$
From the given data, obtain
3r² + 0.16r + 0.0016 = 0

Solve with the quadratic formula.
r = (1/6)[-0.16 +/- √(0.0064)]
r = -0.0133 m, or r = -0.04 m.

Answer:
There are two configurations, as shown in the second figure
(a) q₃ is placed 1.33 cm to the left of q₂.
(b) q₃ is placed 2 cm to the left of q₁.

Answer 2

1) Let's put $q_1$ and $q_2$ both on the x-axis such that $q_2$ is on the right side of $q_1$. The distance between the two charges is $d=2~cm$. 

2) Let's put $q_3$ on the x-axis and let's call $r$ the distance between charge 1 and charge 3. As a consequence, the distance between charge 2 and 3 will be $d+r$: if r is positive, then charge 3 will be located on the left of charge 1, if r is negative, charge 3 will be located between charge 1 and charge 2.

3) The electrostatic force between charge 1 and charge 3 is
$F_{13} = \frac{q_1 q_3}{4 \pi \epsilon_0 r^2}$
while the force between charge 2 and charge 3 is
$F_{23} = \frac{q_2 q_3}{4 \pi \epsilon_0 (r+d)^2}$

4) The problem says the two forces are equal in intensity. Therefore we can write $F_{13} = F_{23}$. Using the equations written at step 3), and substituting $q_1=q$ and $q_2=4q$ as mentioned in the problem, $F_{13} = F_{23}$ becomes
$\frac{q}{r^2}= \frac{4q}{(r+d)^2}$
Using $d=2~cm$, we can solve the equation, and we get two solutions:
$r=2~cm$
$r=-0.67~cm$
Both solutions are correct. With the former, the charge 3 is located 2 cm on the left of charge 1, while with the latter, charge 3 is located 0.67 cm on the right of charge 1, between charge 1 and 2.