E.x. a positively charged metal rod touches a neutral metal rod. Free electrons from the neutral rod will then flow (transfer) to the charged rod, leaving the formerly neutral rod now slightly positively charged
Induction: Charge distribution altered by bringing two objects close, but not touching
Unlike conduction, induction doesn’t alter the net charge of objects when the inducer is taken away
However, induction can redistribute the existing charges on the induced object
Grounded Objects
Objects can be ground to the earth with a conducting wire
The earth is very large and can conduct, so it easily accepts/gives up electrons
Therefore, when an object is induced by another charged object, the original objects will become charged
If the wire is ever cut when the object is under induction, the charge will stay in the object
Electric fields extend outward from every charge and permeates all of space
$$\overrightarrow E = \lim_{q\to0}\frac{\overrightarrow F}{q} \Longrightarrow \overrightarrow F = q \overrightarrow E$$
.$q$ is a positive charge
.$\overrightarrow F$ is the forces the field exserts on .$q$
Has units newtons per coulomb (.$\text{N/C}$)
We can combine this with Coulomb’s law to get
$$\overrightarrow E = \frac{kqQ/r^2}{q} = k \frac{Q}{r^2}$$
We see that .$\overrightarrow E$ is independent of the non-source particle .$q$
.$Q$ is the particle that is responsible for the field in the first place
An electric field at a given point is the sum of all other electric fields that act on that point
$$\overrightarrow E = \overrightarrow E_1 + \overrightarrow E_2 + …$$
21.7 Electric Field Calculations for Continuous Charge Distributions
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We can extend our previous definition to calculus as
$$\overrightarrow E = \int d \overrightarrow E = \int k\frac{1}{r^2}\ dq$$
.$dq = \lambda\ dl \text{ (line)} = \sigma\ dA \text{ (disk)} = \rho\ dV \text{ (sphere)}$
When solving problems, it’s a good idea to use symmetry, check charge direction, and (when applicable) use bounds of .$r \in [0, \infty]$
We can write equation for an infinite plane holding a uniform surface charge density .$\sigma$
$$2A \cdot \overrightarrow E = \frac{\sigma A}{\varepsilon_0} \Longrightarrow \overrightarrow E = \frac{\sigma}{2\varepsilon_0}$$
This also applies in the case where a charge is close to an infinite surface (so that the distance to the surface is much greater than the distance to the edges)
In the case where there are two oppositely charged sheets parallel to one another, the field is .$\vec E = \frac{\sigma}{\varepsilon_0}$ since there are two charges creating the field
The case involving an infinitely long wire can be written generally as
$$\overrightarrow E \cdot 2\pi RL = \frac{\lambda L}{\varepsilon_0} \Longrightarrow \overrightarrow E = \frac{\lambda}{2\pi\varepsilon_0 \cdot r}$$
To visualize electric fields, we draw electric field lines or lines of force
Three properties of Electric Field Lines:
Electric field lines indicate the direction of the electric field; the field points are in the direction tangent to the field line at any point – see point .$P$ in .$\text{(a)}$
The lines are drawn so that the magnitude of the electric field, .$E$, is proportional to the number of lines crossing unit area perpendicular to the lines (i.e. a circle ‘hugging’ a point charge). The closer together the lines, the stronger the field.
Electric field lines start on positive charges and end on negative charges; and the number starting or ending is proportional to the magnitude of the charge.
.$\text{Density} = \frac{\text{number of lines crossing surface}}{\text{area surface}}$
In the case of two oppositely charged parallel & equally spaces plates – such as case .$\text{(d)}$ – we can write the field as
$$\overrightarrow E=\text{const.} = \frac{\sigma}{\varepsilon_0}=\frac{Q}{\varepsilon_0 A}$$
.$Q=\sigma A$ is the charge on one plate of area .$A$
Field lines never cross because it wouldn’t make sense for an electric field to have two directions at the same point.
If .$q_1 q_2 > 0$ (same sign, repulse), then the force and unitary vectors both point away from the two charges
If .$q_1 q_2 < 0$ (opposite sign, attract), then the force vector points towards the two charges and the unitary direction vector still points away from the two charges
Superposition Principle
In a system considering multiple (3+) charges, forces acting on .$q_1$ by .$q_2$ (.$F_{12}$) is independent from whether other charges are present
Total forces acting on .$Q_1$ can be written as .$\overrightarrow F = \overrightarrow F_{12} + \overrightarrow F_{13} + \dots$
Remember to break down the vectors into .$x/y$ components when adding them
E.x. .$F_{1x} = F_{12x} + F_{13x} + \dots$
Realize that the axis are arbitrary
.$\theta = \tan^{-1}\Big(\frac{F_x}{F_y}\Big)$
Charges in Fields
Charge moving with .$\vec v$ that is parallel to uniform field .$\overrightarrow E$
.$\overrightarrow F = q \overrightarrow E = m \vec a \Longrightarrow a_x = \frac{q}{m}\overrightarrow E = \text{const.}$