# 02. Analysis Tools

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$$$\newcommand{\AA}{\unicode[.8,0]{x212B}}$$

# Point Charges

The three charges below are located as shown. Each grid square has width a. Find the net electric force on the positive charge.

pic

To find the electric force on the positive charge, you must first find the electric field at its location in space. Thus, this is really just more practice in calculating the electric field!

The electric field at the location of the positive charge will be the vector sum of the electric field from the -2q charge (E-2Q) and the electric field from the -2q charge (E-2q). Let's calculate these two fields separately and then add them together.

From the -2q charge:

pic

and from the +2q charge:

pic

Adding these two contributions together yields

pic

So the force on the positive charge is:

pic

The force on the positive charge is to the left, and slightly downward, attracted to the two negative charges.

# Force and Motion

In many applications, oppositely charged parallel plates are used to "steer" beams of charged particles. In this example, a proton is injected at 2.0 x 106 m/s into the space between the plates. The plates are 2.0 cm long. What charge density is needed on the plates to give proton a y-velocity of 2.0 x 105 m/s as it exits the plates?

pic

Since this problem involves the motion of a particle between two distinct events, let's complete a motion table.

Event 1: The proton enters the device. Event 2: The proton exits the device.

t1 = 0 s t2 =

r1x = 0 m r2x = 0.02 m

r1y = 0 m r2y =

v1x = 2 x 106 m/s v2x =

v1y = 0 m/s v2y = 2 x 105 m/s

Between the plates, the proton will experience an electrical force, and hence acceleration, in the y-direction. Remember from mechanics that this y-acceleration will not affect thte kinematics of the proton in the x-direction. Thus, in the x-direction the acceleration of the proton is zero.

Applying the kinematic equations in the x-direction yields:

pic

Applying the same kinematics equations in the y-direction yields:

pic

Now, using Newton's Second Law and the relation for electric field from parellel conducting plates, find the necessary charge density.

pic

The bottom plate should be made positive and the top plate negative, both with this charge density. (Note that the force of gravity acting on the proton is completely insignificant compared to the electrical force. This is generally true and we will typically ignore the force of gravity acting on individual particles such as protons and electrons.)

This page titled 02. Analysis Tools is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by Paul D'Alessandris.