# Question about particles in electric fields. Homework question

I am currently working on a homework question and I realized that I don't really understand some of the material that we covered so I hoped that you guys can help me clear up some misunderstandings.

In the diagram I drew (the best I could do with a mouse and paint), there are two positive point charges $Q$ at $y_1=\frac{ab}{2}$ and $y_2=-\frac{ab}{2}$. A microsphere is placed at distance $ab$ from the origin. The microsphere has the charge $-q$ and mass $m$. Note that $q<<Q$

I want to find the equation of motion of the Microsphere and some numerical solutions to this equation.

This is what I got so far:

The two point charges are going to create a field that attracts the microsphere $q$. That field will exert some Force $F$ on $q$.

Since $F=ma$ I can find the equation of motion if I know the force acting on q.

Assuming I know the value of the electrical field $\vec{E}$ at all points then the force acting on a charge $q$ in that field is:

$$\vec{F}=q\vec{E}$$

I also know that because of the superposition principle the electrical field $\vec{E}$ created by two charges is just the vector sum of those two fields.

$$\implies \vec{E}_{total}=\vec{E}_1+\vec{E}_2$$

In addition I know the field created by a single charge is:

$$\vec{E}=k\frac{Q}{(\vec{r})^2}$$

I feel like I have all the information I need to solve this problem but I am somehow too dumb to put all of this together. Can somebody maybe give me a few hints.

• You should try analyzing the force on q on the two axes x and y. You can find the cosine of the angle with of the force with x axon because you know all the sides. – Constantine Black May 7 '15 at 18:05
• But two of the sides are changing right? I want to know $\vec{E}_{total}$ along the x-axis with respect to x but since the two sides are chaning when q is moving I don't know how to figure x out. Maybe my trigonometry is failing me here – qmd May 7 '15 at 18:23
• I understand what you say. What would you say the needed variables are for describing the motion? – Constantine Black May 7 '15 at 18:26
• the vertical field (and therefore force) from the two stationary charges will cancel, so you only have to worry about the horizontal component... – danimal May 7 '15 at 18:27
• Yes my q only moves on the x-axis. That was actually given in the question. – qmd May 7 '15 at 18:29

$$F_{1x}=F_1 cos u$$ where u the angle between the x axon and the distance d between q and Q It is: $$F_{1x}=F_1 cos u ={ k Qq \over d^2}cosu = {k Qq \over \sqrt{x^2 + y_0 ^2}}cosu={k Qq \over \sqrt{x^2 + y_0 ^2}}{\sqrt{x^2 + y_0 ^2} \over x}$$ So, $$F_{1x}={k Qq \over \sqrt{x^2 + y_0 ^2}}{1 \over x}$$ where $y_0$ is the place on y of the Q charge and $$d^2=x^2 + y_0 ^2$$ and $$cosu={d \over x} ={\sqrt{x^2 + y_0 ^2} \over x}$$ The same exactly for the force from $-y_0$. Thus: $$F=2F_{1x}$$
From here you have: $$F=m{d^{2} x \over dt^2}$$ and you may solve the differential equation.