# Intro - Fields of... Electricity?
Ever wondered why, no matter what you do, you just can't seem to put two charged plates together? Ever wondered why, if you put two electrons together, they would drift away from each other?
**Electric fields** are what you're looking for. Defined as the **[[Coulomb's Law (ElectroMagnetism)|Electric Force]] per unit charge due to another charge**, it measures the strength to which another charge exerts a force - a pull - on another. Typically, this charge within the field would be a **positive test charge** - with negligible mass, volume, and for all purposes only carrying a charge of +1.
These 'fields' are **important.** From being the working mechanism for [[Capacitance (ElectroMagnetism)|capacitors]] to helping us understand lightning, electric fields *permit* themselves near-unrivalled importance in our daily lives (no pun intended). So what better thing to do but to learn about them?
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# Millikan's Oil Drop Experiment
# Electric Fields due to Point Charges - Electrifying
Below is a diagram of a positive point charge, with an (imaginary) radius of 0. Meet its' other friend, another positive test charge, say, a distance $r$ away from the other charge.
![[Pasted image 20231102222708.png]]
*An example positive test charge - we'll come back to this later!*
Remember - like charges repel, and so will these! As a result, it's important that we try to model how this can happen and how it happens - or the magnitude, and direction of such a 'force'.
This force is our electric field - a persistent, theoretically infinite source of 'attraction' and 'repulsion', given by the arrows in the diagram above. For electric fields, positive test charges are typically used as a reference - they're portrayed with outwards arrows as above while negative charges typically have the arrows going into them - as shown!
![[Pasted image 20231102224357.png]]
*A negative charge in terms of electric fields.*
Mind you, influence wanes as distance grows, and we can express this relationship with an inverse square relationship. Find the relative strength of a point charges' electric field using this equation!
$E = \frac{F}{q} = \frac{Q}{4\pi \epsilon_{0}r^2}$
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# The Electric Field between two Charged Plates
If you've read [[Capacitance (ElectroMagnetism)|capacitance]] already, chances are that you might know that the **electric field between two charged plates is uniform.** So in other words, if you put a 'charged particle' (say, a droplet of oil) between two oppositely charged plates, the force per unit charge incident on it will be the no matter where you put it!
![[parallelplate-e-field.png]]
*Fig 1: The Uniform Electric Field between two Plates.*
So, when we put a positively charged plate opposite a negatively charged plate, an electric field (as denoted by the white arrows) will be created between the plates.
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To draw this type of field, make sure you remember to:
- Keep the arrows **evenly spaced.**
- The stronger the electric field, the closer together it is.
Also - if we look above to our definition of the electric field, taking the p.d across the two plates we can get a value for the electric field strength:
$E = \frac{F}{q} = \frac{\Delta V}{d}$
Where both of these quantities have $\mathrm{N C^{-1}}$ as their units. Mind you, since the field due to the parallel plates doesn't follow the inverse square law, you need to get familiar with when you can use these equations!
>[!Success]- Movement of Particles within Electric Fields
>The directions of an electric field are given by the arrows - but only for a **positive test charge.** So only positive charges (i.e protons) will follow the vectors given.
>
>So what about electrons? Think [[Power, Work Done and Energy#Mechanical Energy - Conservation of Energy|conservation of energy]]. If it pushes a positive charge - of which the base quantity of 'charge' is the polar opposite to the electron - one way, the electron must therefore **do work against the field if it is within it.** So **negative particles will oppose the direction of the electric field!**
>
>You could probably work this out on your own, but I'm just putting this here as a means of explicit statement /shrug.
# Further - Electric Fields due to Dipoles
Sometimes, we find ourselves with an **electric dipole** on our hands to calculate the electric field at certain points for.
But what IS a dipole? It's a pair of charged particles with an equal magnitude but opposite sign that interact with each other. See the diagram below!
![[dipolediag.png]]
*Fig 1: The Dipole.*
For a dipole, we define an axis that goes through both of the charges - the **dipole axis**, denoted by $z$.
![[dipoleaxis.png]]
*Fig 2: The Dipole Axis*
Now, let's put the point charge ON this dipole axis.
![[chargeanddistancewithdipoleaxis.png]]
*Fig 3: The Dipole, with distance demarcated.*
The distance between the two charges is denoted by "$d
quot;. We know that the electric field of the positive charge points radially outwards, whereas the electric field of the negative charge points radially inwards. Therefore, finding the total electric force is akin to finding the total electrostatic force at the point P, which is given by the equation:
$E_{tot} = E_{+} - E_{-}$
Using [[Coulomb's Law (ElectroMagnetism)|Coulomb's Law]] and the definition of the electric field strength we can write:
$E_{tot} = \frac{Q}{4\pi \epsilon_{0} r_{(+)}^2} - \frac{Q}{4\pi \epsilon_{0} r_{(-)}^2}$
$E_{tot} = \frac{Q}{4\pi \epsilon_{0}} (\frac{1}{r_{(+)}^2} + \frac{1}{ r_{(-)}^2})$
Note the diagrams above. The distance from the positive charge is the distance $z$ minus half the distance $d$, whilst the distance from the positive charge is the distance $z$ PLUS half the distance $d$. So, let's make the substitution.
$E_{tot} = \frac{Q}{4\pi \epsilon_{0}} (\frac{1}{( z-\frac{1}{2}d )^2 } - \frac{1}{( z+\frac{1}{2}d)^2})$
Rewrite for common denominator:
$E_{tot} = \frac{Q}{4\pi \epsilon_{0}} ( \frac{( z-\frac{1}{2}d)^2 - ( z+\frac{1}{2}d)^2}{( z-\frac{1}{2}d )^2 \times( z+\frac{1}{2}d)^2})$
This is extremely tedious.
$E_{tot} = \frac{Q}{(4\pi\epsilon_{0})} \frac{\left( \left( 2zd \right) \right)}{\left( z^2 - zd + \frac{1}{4}d^2 \right)\left( z^2+zd+\frac{1}{4}d^2 \right)}$
Look at the common multiple - the 'z' WILL cancel out. For now, let's rewrite this by factoring out the $z^2$ term, giving us:
$E_{tot} = \frac{Q}{4\pi\epsilon_{0}z^2} \frac{\left( \left( \frac{2d}{z} \right) \right)}{\left( 1 - \frac{d}{z} + \frac{d^2}{4z^2} \right)\left( 1+\frac{d}{z}+\frac{d^2}{4z^2} \right)}$
Now, see where we can make a possible simplification? It's so stupid and simple - non-intuitive to intuitive. It's great! Let's move that 'z' factor from the numerator down and multiply the two square terms to get:
$E_{tot} = \frac{Q}{4\pi \epsilon_{0}z^3} \left( \frac{2d}{\left( 1 - \frac{d^2}{2z^2} + \frac{d^4}{16z^4} \right)} \right)$
Wuh oh! That denominator looks... odd. Watch what happens if we let $(d/2z)^2$ be a variable though!
$E_{tot} = \frac{Q}{4\pi \epsilon_{0}z^3} \left( \frac{2d}{\left( 1 - 2x + x^2 \right)} \right)$
Now THAT'S some monkey business right there! So, rewrite it in terms of $d$ to get:
$E_{tot} = \frac{Q}{4\pi \epsilon_{0}z^3} \left( \frac{2d}{\left( 1 - \left( \frac{d}{2z} \right)^2\right)} \right)$
Time to approximate! Imagine the charge being far, FAR away, so that $\frac{d}{2z}$ is << 1, or much, MUCH less than 1. In other words, $\frac{d}{2z}$ is tiny, so we don't have to factor that in. [^1]
$E_{tot} = \frac{2dQ}{4\pi \epsilon_{0}z^3}$
$E_{tot} = \frac{Qd}{2\pi \epsilon_{0}z^3}$
Since $\frac{1}{2\pi \epsilon_0}$ is a constant, we typically write it as:
$E_{tot} = \frac{1}{2\pi \epsilon_{0}} \frac{Qd}{z^3}$
Where:
- $z$ is the distance from the **dipole centre to the particle**
- $d$ is the **distance between the charges**
- $Q$ is the **charge on the particle being acted on by the net electric field.**
Neat! That equation above is the *general equation for the electric field for a point on the dipole axis.* It's one of two cases you'll need to know, the other one being **the electric field on a point along the perpendicular bisector.**
![[perpendicular bisector.png]]
*Fig 4: Now the charge is on the perpendicular bisector!*
Now, for this case, note the directions of the electric fields.
![[net forces on perpendicular bisector charge.png]]
*Fig 5: Net forces!*
Here, the net force due to both of these charges can be written as $2E\sin\theta$ since it is obvious that the horizontal components of the two forces actually cancels out. Therefore:
$E_{tot} = 2\left( \frac{Q}{4\pi \epsilon_{0}r^2} \right) \sin \theta$
Where $r$ is the distance from each charge in the dipole to the charge within the field. Now, we need to find an expression for $sin \theta$, being:
$\sin \theta = \frac{d}{r}$
Therefore:
$E_{tot} = \frac{Qd}{2\pi \epsilon_{0}r^3}$
[^1]: Check out question 22.19 for cases where the ratio $\frac{d}{2z}$ isn't negligible!
# Example Questions
*Textbook: Halliday, Resnick, Walker - 10th Edition!*
>[!Example]- Chapter 22 Problem 20
>
>**Question:**
>
>Consider a point P on that axis at distance $z = 5.00d$ from the dipole center ($d$ is the separation distance between the particles of the dipole). Let $E_{appr}$ be the magnitude of the field at point P as approximated by Eqs. 22-8 and 22-9. Let $E_{act}$ be the actual magnitude. What is the ratio $E_{appr}$/$E_{act}$?
>
>**Answer:**
>Let's find $E_{appr}$ as a function first, from the final equation we get:
>$\frac{1}{2\pi \epsilon_{0}} \left( \frac{qd}{z^3} \right)$
>$\frac{1}{2\pi \epsilon_{0}} \left( \frac{q}{125} \right)$
>
>Then, let's find $E_{act}$, which is the total electric field $E_{+} - E_{-}$:
>$E_{act} = \frac{Q}{4\pi \epsilon_{0}(4.5)^2} - \frac{Q}{4\pi \epsilon_{0}(5.5)^2}$
>So to find the ratio, we can just divide:
>$\frac{E_{appr}}{E_{}}\frac{\frac{1}{2\pi \epsilon_{0}} \left( \frac{q}{125} \right)}{E_{act} = \frac{Q}{4\pi \epsilon_{0}(4.5)^2} - \frac{Q}{4\pi \epsilon_{0}(5.5)^2}}$
>$-\frac{\frac{1}{250}}{4.0 * 10^{-3}} = 0.9801$
>[!Danger]- Chapter 22 Problem 21
>_Electric quadrupole. Figure 22-46 shows a generic electric quadrupole. It consists of two dipoles_ with dipole moments that are equal in magnitude but opposite in direction. Show that the value of E on the axis of the quadrupole for a point P a distance z from its center (assume $z \gg d$) is given by:
>
>$E_{tot} = \frac{3Q}{4\pi \epsilon_{0}}$
>in which Q ( $2qd__2__) is known_
# What's Next?
Nice! Of course, there's still more to consider about electric fields - such as the electric field due to a line or a 'coil' - a square-shaped circuit - but this is all the important stuff, for the most part. Try to derive those yourself - very fun and offers great integral practice.
Try taking a look at [[Electric Potential (ElectroMagnetism)|Electric Potential]] next!