# a2 physics

this is what i know for capacitors, C = Q / V, they store charge, how???

i know these equations E = 1/2QV C = C1 + c2 etc

a i) A capacity has A = 400cm2, d = 2.00 mm, and air ( er = 1.00 ) as dielectric. Its plates are connected to a battery of 12.0V so that it is fully charged.

Calculate the energy stored in the capacitor

ii) The capacity is then disconnected from the battery, the charges on its plates remaining unaltered, and isolated. A sheet of class Er = 7.5 is inserted so that it fills the space between the plates exactly. Calculate the energy stord in the capacitor

iii) Account for the difference between your answers in b1 and b2
BloBoy
this is what i know for capacitors, C = Q / V, they store charge, how???

i know these equations E = 1/2QV C = C1 + c2 etc

a i) A capacity has A = 400cm2, d = 2.00 mm, and air ( er = 1.00 ) as dielectric. Its plates are connected to a battery of 12.0V so that it is fully charged.

Calculate the energy stored in the capacitor

ii) The capacity is then disconnected from the battery, the charges on its plates remaining unaltered, and isolated. A sheet of class Er = 7.5 is inserted so that it fills the space between the plates exactly. Calculate the energy stord in the capacitor

iii) Account for the difference between your answers in b1 and b2

capacitors store charge and energy by holding opposite charges on their plates.

for a1) ummmmm now this is tricky because you have to use two equations in one

you are right in saying w = 1/2 QV, but there is another equation W = 1/2CV^2

W = 1/2 CV^2

now for C, the capacitance we can use another equation which is AEoEr/d
Eo is the permittivity of free space and Er is the permittivity of the medium
= 1/2 x AeoEr/d V^2

= 1/2 x 400 x 10^-4 x 8.8 x 10^-12 x 1 x 12^2 / 2 x 10^-3

= 1.27 x 10^-8 J

a2) Divide the energy stored ( 1.27 x 10^-8 ) by the relative permittivty (7.5 )

W = 1.27 x 10^-8 / 7.5 = 1.7 x 10^-9

and for part 3........ ummmm you could say that the glass molecules are polarized so when a sheet of glass is inserted, more energy is stored? or when the diaelectric is inserted into the capacitor..... each plate induces an opposite charge on the surface near it

i'm not sure for part 3.......... some1 plz check my math for parts 1 and 2

hope i helped
They store charge by being connected up to a battery, because they are effectively two metal plates seperated by a small layer of air, charge can't flow, but it accumulates on the metal plates, positive charge on one, and negative on the other. Now if you disconnect it from a battery, the charge has to stay where it is, so you have effectively stored charge on the plates.

a i) A capacitor has A = 400cm2, d = 2.00 mm, and air ( er = 1.00 ) as dielectric. Its plates are connected to a battery of 12.0V so that it is fully charged.

What this means is that the area of the metal plates is 400 square centimetres, they are seperated by 2mm and you can assume that the plates are in vacuum, strictly speaking the relative permmitivity of air is a bit more than 1, but we normally assume it is 1.

Now to work out the capacitance of a parallel plate capacitor you use the equation

C=ere0A/d

where e0 is the permittivity of free space and is equal to 8.85x10^-12 F/m

So C = (8.85*10^-12)*0.04/0.002 = 1.77*10^-10 F

We also know that C = Q/V and E = 1/2QV

Combinding that we see that E = 1/2CV^2

So the energy stored in this capacitor when it is connected to a 12V power supply is

E = (1/2)*(1.77*10^-10)*12^2 = 1.27*10^-8 J

But I do not understand the rest of the question, inserting a piece of glass into a charged capacitor will change the capacitance, but it will not alter the amount of charge, or the amount of energy stored in it. It would alter those properties if it was charged again, so are you sure that you wrote the question out correctly?
am glad i got same answer ......... most of the letters/numbers on my calculator are worn off :/
I don't understand why you said that the energy falls when a piece of glass is put in the capacitor though. The higher relative permitivity increases Capacitance, but once the Capacitor is removed from the circuit surely it is a closed system and the energy cannot change?
AntiMagicMan
I don't understand why you said that the energy falls when a piece of glass is put in the capacitor though. The higher relative permitivity increases Capacitance, but once the Capacitor is removed from the circuit surely it is a closed system and the energy cannot change?

ok going back to what i said earlier........ as the dielectric is inserted/put into the capacitor........ each plate induces an opposite charge on the dielectric surface that is near it.... the dielectric is therefore attracted INTO the capacitor and since the system does work on the dielectric the capacitor loses energy??
AntiMagicMan

C=ere0A/d

where e0 is the permittivity of free space and is equal to 8.85x10^-12 F/m

just wondering how many constants we have to learn for as (and a2) physics?
MC REN
just wondering how many constants we have to learn for as (and a2) physics?

The constants should be given to you on a data sheet?

It's strange but I don't recall learning about this from A2! I should remember it . I only remember e0 being used as part of the k constant in the force between two charged particles. I must have bad memory!
LifeWired
The constants should be given to you on a data sheet?

not all
DazYaABBB
not all

Apparently I survived then
LifeWired
The constants should be given to you on a data sheet?

It's strange but I don't recall learning about this from A2! I should remember it . I only remember e0 being used as part of the k constant in the force between two charged particles. I must have bad memory!

I only did Q = CV and and the parallel and series equations for capacitance at A2 (Edexcel). The stuff I know about dielectrics and working out C for a given configuration of capacitor is from the first year of my degree.

As for the constants, you don't need to learn any in my experience. At A-level you get given them, and for my degree you are given the constants needed on the exam booklet.

They are testing your knowledge of physics, not your ability to remember arbitrary numbers. They can however ask you questions about the constants, i.e. what does the permitivity of freespace represent? (Disclaimer: Don't know if they would ask that at A-level.) so you still have to know about them.
This is reassuring, I'll be doing a lot of this in the future