Three charged particles lie in the $xy$-plane at an angle $\theta $ from the $x$-axis. The charges each have magnitude $q$ $\left(q > 0\right)$, but $q_1=q_2=+q$ and $q_3=-q$. Charge $q_1$ is located at the origin, charge $q_3$ is a distance $3r$ from $q_1$, and both of them are fixed. Charges $q_1$ and $q_2$ are connected by an ideal neutral spring of constant $k_s$, and therefore $q_2$ can move. The spring is initially neither compressed nor stretched, and $q_2$ is initially a distance $r$ from $q_1$.

1. Express the net electric force (in vector form) initially acting on charge $q_2$.

View answer

Initially, the spring is neither stretched nor compressed and therefore there is no spring force (until $q_2$ moves and stretches the spring).

The charge $q_1$ is a distance $r$ from charge $q_2$ and exerts a repulsive force with magnitude $F_{1\ on\ 2}$ equal to

F_{1\ on\ 2}=\frac{k\left|q_1q_2\right|}{r^2}=\frac{kq^2}{r^2}

The charge $q_3$ is a distance $2r$ from charge $q_2$ and exerts an attractive force with magnitude $F_{3\ on\ 2}$ equal to

F_{3\ on\ 2}=\frac{k\left|q_2q_3\right|}{{\left(2r\right)}^2}=\frac{kq^2}{4r^2}

The net force acting on $q_2$ is therefore given by

\begin{aligned} {\overrightarrow{F}}_{net}&=\left[F_{1\ on\ 2}{\mathrm{cos} \left(\theta \right)\ }+F_{3\ on\ 2}{\mathrm{cos} \left(\theta \right)\ }\right]\hat{i}+\left[F_{1\ on\ 2}{\mathrm{sin} \left(\theta \right)\ }+F_{3\ on\ 2}{\mathrm{sin} \left(\theta \right)\ }\right]\hat{j} \\ \\ &=\left[\frac{kq^2}{r^2}+\frac{kq^2}{4r^2}\right]{\mathrm{cos} \left(\theta \right)\ }\hat{i}+\left[\frac{kq^2}{r^2}+\frac{kq^2}{4r^2}\right]{\mathrm{sin} \left(\theta \right)\ }\hat{j} \\ \\ &=\frac{5kq^2}{4r^2}{\mathrm{cos} \left(\theta \right)\ }\hat{i}+\frac{5kq^2}{4r^2}{\mathrm{sin} \left(\theta \right)\ }\hat{j} \end{aligned}

The net force acting on the charge $q_2$ initially is equal to

\boxed{{\overrightarrow{F}}_{net}=\frac{5kq^2}{4r^2}{\mathrm{cos} \left(\theta \right)\ }\hat{i}+\frac{5kq^2}{4r^2}{\mathrm{sin} \left(\theta \right)\ }\hat{j}}

2. Express the spring force (as a vector) acting on charge $q_2$ after a small displacement of magnitude $\mathrm{\Delta }r$. Explain in words the direction of the force.

View answer

The charge $q_2$ will move away from charge $q_1$ and stretch the spring by an amount $\mathrm{\Delta }r$. The spring force will therefore be directed toward $q_1$ and have a magnitude $F_{spring}$ equal to

F_{spring}=k_s\mathrm{\Delta }r

Thus, in vector form, the spring force acting on charge $q_2$ is equal to

\boxed{{\overrightarrow{F}}_{spring}=-k_s\mathrm{\Delta }r{\mathrm{cos} \left(\theta \right)\ }\hat{i}-k_s\mathrm{\Delta }r{\mathrm{sin} \left(\theta \right)\ }\hat{j}}

3. Express each of the forces, in vector form, once $q_2$ has reached its equilibrium position at distance $r_e$ from charge $q_1$.

View answer

Once the charge $q_2$ has reached its equilibrium position, it will have stretched the spring an amount $\mathrm{\Delta }r=r_e-r$ and the spring force acting on it will be equal to

\boxed{{\overrightarrow{F}}_{spring}=-k_s\left(r_e-r\right){\mathrm{cos} \left(\theta \right)\ }\hat{i}-k_s\mathrm{\Delta }\left(r_e-r\right){\mathrm{sin} \left(\theta \right)\ }\hat{j}}

Once the charge $q_2$ has reached its equilibrium position, it will be a dstiance $r_e$ from charge $q_1$ and a distance $2r-r_e$ from charge $q_3$. The force from $q_1$ will then be equal to

\boxed{{\overrightarrow{F}}_1=\frac{kq^2}{r^2_e}{\mathrm{cos} \left(\theta \right)\ }\hat{i}+\frac{kq^2}{4r^2_e}{\mathrm{sin} \left(\theta \right)\ }\hat{j}}

and the force from $q_3$ will be equal to

\boxed{{\overrightarrow{F}}_3=\frac{kq^2}{{\left(2r-r_e\right)}^2}{\mathrm{cos} \left(\theta \right)\ }\hat{i}+\frac{kq^2}{{\left(2r-r_e\right)}^2}{\mathrm{sin} \left(\theta \right)\ }\hat{j}}

4. Write the scalar equation satisfied by $r_e$ stating that $q_2$ is in static equilibrium. No need to simplify.

View answer

We apply the Equilibrium Condition to the charge $q_2$ which yields the following equation

\begin{aligned} \left\{ \begin{array}{c} F_{net\ x}=0 \\ \\ F_{net\ y}=0 \end{array} \right.\ \ \ \ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c} \displaystyle{-k_s\left(r_e-r\right){\mathrm{cos} \left(\theta \right)\ }+\frac{kq^2}{r^2_e}{\mathrm{cos} \left(\theta \right)\ }+\frac{kq^2}{{\left(2r-r_e\right)}^2}{\mathrm{cos} \left(\theta \right)\ }=0} \\ \\ \displaystyle{-k_s\mathrm{\Delta }\left(r_e-r\right){\mathrm{sin} \left(\theta \right)\ }+\frac{kq^2}{4r^2_e}{\mathrm{sin} \left(\theta \right)\ }+\frac{kq^2}{{\left(2r-r_e\right)}^2}{\mathrm{sin} \left(\theta \right)\ }=0} \end{array} \right. \\ \\ &\Rightarrow \ \ \ \ \ \boxed{-k_s\mathrm{\Delta }\left(r_e-r\right)+\frac{kq^2}{4r^2_e}+\frac{kq^2}{{\left(2r-r_e\right)}^2}=0} \end{aligned}

Note: the two equations simplify to the same equation because there is only one single direction along which all the forces act. Effectively, you end up with the equation you would have derived if you have written he equilibrium condition directly along the direction of the forces.

Previous Topic

Back to Lesson

Next Topic