A Capacitor is a device that stores electric charges. It has two metals being separated by an insulator "Dielectric" to prevent discharge. To discharge a Capacitor, you join the two metals together. Get a circuit board(looks like green board), say from your charger, notice a device that has two metals on same position, usually looks like a Battery/Drum Shape, that's a Capacitor. also notice a rubber like material on the top of the Capacitor where the two metals are fixed, that's the Dieletric. Note, a Battery is different from a Capacitor. A Battery supplies electric charges while Capacitor stores electric charges.
Symbol of Capacitor -I I- the vertical lines will be equal in length.
Symbol of a Cell/Battery -I i- the vertical lines are not equal, one will be higher than the other in length.
A Capacitance of a Capacitor is the ability of the Capacitor to store electric charges. It's unit is farads(F) or microfarads(;uF).
Continue to read below....

The formular to find the Capacitance of a Capacitor is given as>
C= Q/V
where C is the capacitance in farads or microfarad; Q is the electric charge in coulombs; V is the voltage in volts.
ARRANGEMENT OF CAPACITORS:
Capacitors are arranged in plates of (a) parallel (b) series.
A) PARALLEL CAPACITORS: They are kept parallel to eachother. That's when you see the symbol up and another down, mostly front and back ways. The formular is
C = C1+C2+C3... that's sum up Capacitors number 1,2,3,....
B) SERIES CAPACITORS: They are kept sideways. Just like your friend sitting next to you in class. The formular>
1/C= 1/C1 + 1/C2 + 1/C3 ... sum up the inverse of Capacitors number 1,2,3...
ENERGY STORED IN A CAPACITOR.
It's given as
W= 1/2 *V *Q
OR
W=1/2 * (Q*Q)/C
OR
W= 1/2 *C *V*V.
where W is workdone to get the energy stored in joules; V is voltage in volts; Q is electric charge in coulombs, C is capacitance in microfarad.
Continue to read below...

rusher14:
Ratio of charge to potential.

Yes, it can be defined as ratio of electric charge,q, to potential difference, v. That's in mathematical statement, no wonder students find it difficult to understand physics with such definition.

APPLICATIONS/USES OF CAPACITORS:
It means where you can make use of Capacitors, such as
1. ignition system of a car.
2. radio circuits for tuning, etc.
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OP, please keep it up! But could u pls give some examples that'll help us apply the equations above? I think understanding Physics is easier when students learn how/when some equations are applied.

Laive:
OP, please keep it up! But could u pls give some examples that'll help us apply the equations above? I think understanding Physics is easier when students learn how/when some equations are applied.

Thanks. Yes, I will. Keep following...

E.g1) a capacitor of charge, 5x10raise_to_power minus 8 coulomb, has a potential of 100v. What is the capacitance of the capacitor?
Soln:
using formular for capacitance
C=Q/V; Where C=?; Q= 5x10'-8 coulombs; V= 100v.
C= 5x10'-8/100
C= 5x10'-10F = 5x10'-4 microfarad.
Nb: i'm using '- to denote "raise to power minus."

E.g2) what is the equivalent capacitance of a 3microfarad and a 6microfarad capacitor connected in parallel?
Soln:
Capacitors in parallel formular
C= C1 + C2;
C=?; C1=3; C2=6
C= 3+6= 9microfarad

E.g3) what is the effective capacitance of 5microfarad capacitor and a 3microfarad capacitor connected in series?
Soln:
Using series capacitance formular
1/C = 1/C1 + 1/C2
1/C = 1/5 + 1/3 = 3+5/15 = 8/15
1\C = 8/15
the C will move up and 1 down, same goes for the R.H.S, 15 will move up and 8 down.
C/1 = 15/8
C = 1.88= approx. 2
C = 2microfarad

Nice

Nice one there! And thanks for making it clear... Any examples on 'Energy stored in a capacitor' ??

Laive:
Nice one there! And thanks for making it clear... Any examples on 'Energy stored in a capacitor' ??

Question: Three capacitors of capacitance, 2μF, 4μF and 8μF are connected in parallel and a p. d of 6V is maintained across each capacitor, what is the total energy stored?

Solution:

Capacitors in parallel, C=C1+C2+C3
=2+4+8
=14μF
Energy stored in a capacitor, E= ½CV²
=½(14x10⁻⁶x6²)
=2.52x10⁻⁴J

Note: micro(μ)=10⁻⁶

well done original poster

E.g4) calculate the energy stored in a 20microfarad capacitor if the p.d between the plates is 40v.
Soln:
W=?; C= 20microfarad; V= 40v.
hint: use the formular that only one quantity is missing but the others given. That's
W= 1/2 x C x V'2
W= 1/2 x (20x10'-6) x 40'2
W= 1.6x10'-2J
Nb: '- denotes raise to power minus. I'm using a java phone to type, sorry!