Problem: Two charged particles, with charges q1 = q and q2 = 4q, are located at a distance d = 2.00cm 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 = 2.00 nC.  Assuming charge 1 is located at the origin of the x-axis and the positive x-axis points to the right, find the two possible values, x3,r and x3,l for the position of charge 3.

FREE Expert Solution

Coulomb's law for point charges::

$\boxed{\mathbf{F}\mathbf{=}\frac{\mathbf{k}{\mathbf{q}}_{\mathbf{1}}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{r}}^{\mathbf{2}}}}$

Quadratic formula:

$\boxed{\mathbf{x}\mathbf{=}\frac{\mathbf{-}\mathbf{b}\mathbf{\pm }\sqrt{{\mathbf{b}}^{\mathbf{2}}\mathbf{-}\mathbf{4}\mathbf{a}\mathbf{c}}}{\mathbf{2}\mathbf{a}}}$

Expansion formular:

$\boxed{{\mathbf{(}\mathbf{a}\mathbf{-}\mathbf{b}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathbf{a}}^2\mathbf{-}\mathbf{2}\mathbf{a}\mathbf{b}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}}$

d = 2.00 cm(1m/100cm) = 0.02 m

Assume the third charge is located between the two charges such that:

r₁₃ = (d - x) m

r₂₃ = x m

q₁ = q = 2.00 nC(10^-9C/1nC) = 2.00 × 10^-9 C

q₂ = 4q = 8.00 nC(10^-9C/1nC) = 8.00 × 10^-9 C

q₃ = q = 2.00 nC(10^-9C/1nC) = 2.00 × 10^-9 C

We are told that F₁₃ = F₂₃:

$\begin{array}{rcl}{\mathbf{F}}_{\mathbf{13}}& \mathbf{=}& {\mathbf{F}}_{\mathbf{23}}\\ \frac{\cancel{\mathbf{k}}{\mathbf{q}}_{\mathbf{1}}\cancel{{\mathbf{q}}_{\mathbf{3}}}}{{\mathbf{\left(}{\mathbf{r}}_{\mathbf{13}}\mathbf{\right)}}^{\mathbf{2}}}& \mathbf{=}& \frac{\cancel{\mathbf{k}}{\mathbf{q}}_{\mathbf{2}}\cancel{{\mathbf{q}}_{\mathbf{3}}}}{{\mathbf{\left(}{\mathbf{r}}_{\mathbf{23}}\mathbf{\right)}}^{\mathbf{2}}}\\ \frac{\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{\times }\cancel{{\mathbf{10}}^{\mathbf{-}\mathbf{9}}}}{{\mathbf{(}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{-}\mathbf{x}\mathbf{)}}^{\mathbf{2}}}& \mathbf{=}& \frac{\mathbf{8}\mathbf{.}\mathbf{00}\mathbf{\times }\cancel{{\mathbf{10}}^{\mathbf{-}\mathbf{9}}}}{{\mathbf{x}}^{\mathbf{2}}}\\ \mathbf{2}{\mathbf{x}}^{\mathbf{2}}& \mathbf{=}& \mathbf{8}\mathbf{(}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{04}\mathbf{x}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{0004}\mathbf{)}\\ \mathbf{0}& \mathbf{=}& \mathbf{8}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{2}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{32}\mathbf{x}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{0032}\\ & \mathbf{=}& \mathbf{6}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{32}\mathbf{x}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{0032}\end{array}$