Simple Charge & Field Examples

Field fsrom several simple point charges

What is the field strength in N/C at $Q_A$?

From Electrostatics:

$$|E|=k_e\frac{Q}{r^2}$$ $$...\overrightarrow{E_{Q_A}}=k_e\left(\begin{array}{c}\left[\begin{array}{c}0\\ \frac{Q_C}{r^2}\end{array}\right]+\left[\begin{array}{c}\frac{Q_B}{r^2}\cos (30)\\ \frac{Q_B}{r^2}\sin (-30)\end{array}\right]\end{array}\right)$$ $$...\overrightarrow{E_{Q_A}}=8.9875\cdot 10^9\cdot \frac{10^{-6}}{r^2}\left(\begin{array}{c}\left[\begin{array}{c}0\\ Q_{C}10^6\end{array}\right]+\left[\begin{array}{c}{Q}_{B}10^6\frac{\sqrt{3}}{2}\\ Q_{B}10^6\frac{1}{2}\end{array}\right]\end{array}\right)$$ $$...\overrightarrow{E_{Q_A}}=8.9875\cdot 10^9\cdot \frac{10^{-6}}{0.222^2}\left(\begin{array}{c}\left[\begin{array}{c}0\\ -4.91\end{array}\right]+\left[\begin{array}{c}10.43\frac{\sqrt{3}}{2}\\ 10.43\frac{1}{2}\end{array}\right]\end{array}\right)$$ $$...\overrightarrow{E_{Q_A}}=\left[\begin{array}{c}9.03\\ 0.305\end{array}\right]$$ $$...|E_{Q_A}|=182361.41547\cdot \sqrt{9.03^2+0.305^2}=1,647,662\frac{\mathbf{N}}{\mathbf{C}}$$

Find the direction of the electric field at the location of $Q_A$ in degrees clockwise from the negative x (horizontal in image) direction.

$$\overrightarrow{E_{Q_A}}=\left[\begin{array}{c}9.03\\ 0.305\end{array}\right]$$ \theta_\vec{{E_{Q_A}}} = \tan^{-1}(\frac{0.305}{9.03}) = 1.934

From the negative horizontal axis, that’s $180-30+1.934=$ degrees.

What is the magnitude, in newtons, of the force on charge $Q_A$ given that $Q_A=1.59nC$?

From Electrostatics:

$$\vec{F}=Q_{Object}{\vec{E}}_{Source}$$ $$|F|=1.59\cdot 10^{-9}\cdot 1,647,662=0.00261978258$$

 Consider a particle of charge $Q=1.1C$ and mass $m=1.2kg$ passing through the region between a pair of infinitely long horizontal plates separated by a distance $d=3.5m$ with a uniform electric field strength $E=28\frac{N}{C}$ directed in the downwards direction (-y direction). The particle begins moving horizontally with an initial velocity of v = 25 m/s from a position halfway between the plates.

How far horizontally, in meters, will the particle travel before striking one of the plates?

From Electrostatics:

$$\vec{F}=Q_{Object}{\vec{E}}_{Source}$$ $$\vec{F}=1.1\cdot 28=30.8N$$

From 5th grade and Kinematics:

$$F=mA$$ $$x=x_0+V_{0x}t+0.5A_x{t}^2$$ $$1.75=0.5(\frac{30.8}{1.2}+9.81)t^2\to t=0.314s$$

Into the same equation…

$$x=25(0.369)=7.85m$$

Calculate the speed, in meters per second, with which the particle will strike the plate.

From Kinematics:

$$V_x=V_{0x}+A_{x}t$$ $$V_x=(\frac{30.8}{1.2}+9.81)0.314=11.139\frac{m}{s}$$ $$\sqrt{11.139^2+25^2}=27.3692769543$$

Suppose so the electric field is directed upward instead of downward. Calculate the new horizontal distance, in meters, that the particle travels before striking one of the plates.

$$1.75=0.5(\frac{30.8}{1.2}-9.81)t^2\to t=0.469s$$ $$25(0.469)=11.725m$$

Square of charges, POI in center

Find the magnitude of the net Coulomb force, in newtons, on the charge $q$ in the figure

$$\overrightarrow{\Sigma E_Q}=\frac{k_e}{r^2}\left(\begin{array}{c}\overrightarrow{Q_A}+\overrightarrow{Q_B}+\overrightarrow{Q_C}+\overrightarrow{Q_D}\end{array}\right)$$

Okay, this is ugly because the vectors are directioned but not magnituded by the distance and now we’re in land of no (reasonable) units… (Actually the units of the vectors is $|Q|\hat{Q}$… so yeah: wack)

$$\overrightarrow{\Sigma E_Q}=\frac{k_e\cdot 10^{-6}}{r^2}\left(\begin{array}{c}\left[\begin{array}{c}-2.8\\ 0\end{array}\right]+\left[\begin{array}{c}0\\ -3.8\end{array}\right]+\left[\begin{array}{c}-4.6\\ 0\end{array}\right]+\left[\begin{array}{c}0\\ -1.1\end{array}\right]\end{array}\right)$$ $$\overrightarrow{\Sigma E_Q}=\frac{k_e\cdot 10^{-6}}{r^2}\left[\begin{array}{c}-7.4\\ -4.9\end{array}\right]$$ $$\overrightarrow{\Sigma E_Q}=\frac{8.9875\cdot 10^3}{0.231}\left[\begin{array}{c}-7.4\\ -4.9\end{array}\right]$$ $$\overrightarrow{\Sigma E_Q}=\left[\begin{array}{c}-287911.255\\ -190643.939\end{array}\right]$$ $$|\Sigma E_Q|=\sqrt{287911.255^2+190643.939^2}=345308.560903$$ $$F=1.1\cdot 10^{-6}\cdot 345308.561=0.3798$$

This is incorrect and I do not know why.