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) = (q_(1)q_(3))/(4 \pi \epsilon_(0)(d+r)^(2)) = (q^(2))/(4 \pi \epsilon_(0)(d+r)^(2))`

`F_(23) = ((4q)q)/(4 \pi \epsilon_(0)r^(2))`

Because F₁₃ = F₂₃, therefore

`(1)/((d+r)^(2)) = (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) = (q_1 q_3)/(4 \pi \epsilon_0 r^2)`
while the force between charge 2 and charge 3 is

`F_(23) = (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

`(q)/(r^2)= (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.