Ch.1 Electric Charges and Force

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Charge

Coulomb's Law

Observation

Force exerted by particle Q1Q_1 with charge q1q_1 on Q2Q_2 with charge q2q_2 a distance of rr away in the direction of u⃗\vec{u} is
F⃗Q1→Q2=Kq1q2r2u⃗\vec{F}_{Q_1\rightarrow Q_2}=K\frac{q_1q_2}{r^2}\vec{u}
This requires no analysis as to positive/negative.


Electric Fields

If one particle moves, does the other particle respond to the change in force immediately?
It doesn't, especially considering if the particles were thousands of lightyears apart.
To determine this, Faraday invented fields
From a Newtonian perspective, particle A directly responds to particle B.
From Faraday's view, particle A affects the space around it and particle B responds to the change in space.
field: in math, a function that assigns a vector to every point in space; in physics, the idea that a physical entity exists everywhere in space

Faraday Model

Postulates:

If a probe chargeqq is placed on the field at (x,y,z)(x,y,z) and force F⃗on q\vec{F}_{\text{on }q} is measured, we can find the electric field at (x,y,z)(x,y,z) by
E⃗(x,y,z)=F⃗on q at (x,y,z)q\vec{E}(x,y,z)=\frac{\vec{F}_{\text{on }q}\text{ at }(x,y,z)}{q}

Field Model

Electric Field of a Point Charge

We use q′q' to probe the electric field of qq.
Define r⃗\vec{r} to be the unit vector from the point of origin pointing toward the point of interest; i.e. the direction vector
By Coulomb's Law, we calculate the force on q′q' as
F⃗on q′=14πϵ0qq′r2r⃗\vec{F}_{\text{on }q'}=\frac{1}{4\pi\epsilon_0}\frac{qq'}{r^2}\vec{r}
It is customary to use 1/4πϵ01/4\pi\epsilon_0 instead of KK for field calculations. Then by the electric field equation, we can find
E⃗=F⃗on q′q′=14πϵ0qr2r⃗\vec{E}=\frac{\vec{F}_{\text{on }q'}}{q'}=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\vec{r}
Calculating the field at several places lets us draw a field diagram.

Field Diagrams

Positive Charge

Negative Charge