Electrostatics
You will start with Coulomb's law concerning the force of electric attraction on point charges q1 and q2 in Coulombs (C) separated by distance r,
F = kq1q2/r2 (where k = 1/4peo = 8.99 x 109 Nm2/C2),
and whatever you do, never plug in numbers and get an answer without making a sketch and considering the direction of forces or fields from the sketch. (Well, you might get by without a sketch for voltage, [a scalar, not a vector], since V gets more + as you go closer to a positive charge and - as you go closer to a negative, and so you can just plug into the equation, V = kq/r for V at distance r from a point charge q. And for more than one charge, V = S kq/r.)
For example, if A, B and C are charges in the sketch below, think about the direction of force on each charge, given that A,B,C are a) all +; b) all -; c) +,+,-; d) +,-,-….(etc.)
Now AB is the length of the segment from the center of A to the center of B. Given that AB = 0.30 m, BC = 0.40 m, AC = 0.50 m, and charge A = 2.0 mC, B = -6.0 mC, and C = -7.0 mC, find the magnitude and direction of the force on each. Note that they form a right triangle.
The above problem is too boring, but it is a test of your understanding of Coulomb's law, which is fundamental. Another thing you should be able to do is calculate the electric field at any point in the picture above. For example, at the midpoint of BC.
Electric field is the net electric force on a tiny charge qo divided by qo. Then it is independent of qo. Show that E = S kq/r2 and be careful to do a vector sum. Imagine a positive qo located where you want to find E, and it will have force away from positive charges and toward negative. Choose an appropriate coordinate system.
Later you will find potential or voltage, an easier thing to deal with, but a harder concept, in a way. V is simply energy/charge, but what energy? What charge? You'd better really learn this. Electricians don't really need to know exactly what V is; some kind of gut feeling about electrical pressure will do. You should know it at a deeper level. Also, know the relationship between voltage and electric field. (If you travel in the direction of E, there is a drop in voltage equal to E times distance traveled. And if E changes, use the average E.)
Finding F, E, or V due to point charges involve summations; if charge is smeared out, the corresponding problem is a sum of infinitely many terms, in other words the integral in calculus. Then there is Gauss's law, which in its most general form is written as an integral, but in most applications, we avoid the integration. Gauss's law relates E to charge and distance, hence it is an alternative to Coulomb's law. It can be regarded as a reformulation of Coulomb's law. You probably do not want to go beyond this point if your text does not deal with Gauss's law.
To understand Gauss's law, you need to know about electric flux. The flux through area A is defined as the electric field component perpendicular to A times A. Thus if we have field E at area A, the absolute value of the flux could be anything from 0 to EA, depending on the direction of E. Use F for flux, so F = E^ A.
Flux is defined to be positive if E is directed out of a closed surface. Gauss's law states that e o times the total flux through a closed surface equals the charge inside (e o = 1/(4p k) = 8.85 x 10-12 C2/(Nm2)). If you consider a point charge or a sphere of charge, and let the "Gaussian surface" be a concentric sphere surrounding it, area 4p r2, and you can easily show that E obtained by Gauss's law is the same as what you get from Coulomb's law. Now consider a long uniformly charged rod of length L, and let's say you want the field near the side of it. With Coulomb's law this involves doing an integral, but with Gauss's law, it is quicker. Consider an imaginary cylinder concentric with the rod, and neglect flux coming out the ends. The area is 2p rL, so eoE2p rL = qinside, and you can solve for E.
If a problem has charge with spherical or cylindrical symmetry, you should probably use Gauss's law in a manner similar to the above discussion. Other doable problems using Gauss's law include finding the field due to a flat surface: If s is charge/area, we find that E = s /eo near a conductor and s /2eo near a non-conductor. (Use an imaginary box with surfaces parallel and perpendicular to the surface to enclose area A. The perpendicular ones have no flux going through, so it is a simple case.)
After you play with the above ideas for a while, you will get a charge out of capacitors. Or maybe you want to zap back to the main page on electricity & stuff.
Or go to one of the other main physics pages:
Mechanics
Vibrations and waves
Quantum
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