Capacitance units help

Hi,

I'm just doing some summary questions from the textbook. For the last question I got the right values for the answers but the wrong order of magnitude and I'm not sure what I'm doing wrong. I feel like it's to do with units but I've got the other questions right using the same method.

A capacitor is charged by using a constant current of 24microamps to a pd of 4.2 in 38s. It is then charged from 4.2V by using a current of 14microamps in 50s.

i. Charged stored at a pd of 4.2
Q=It = 24×38= 912micro coulombs (answer given in the back as 0.91 micro coulombs)

ii. Capacitance = Q/V= 0.91/4.2 = 0.22 micro Farads. (Correct)

iii. The extra charge stored at a current of 14 micro amps.
Q=It= 14×50= 700 micro coulombs (correct)

iv. The pd after the extra charge was stored
V= Q/C = 700/0.22 = 3200 which should be 3.2 as the final answer is 3.2+4.2=7.4

Any help would be really apreciated

Remember it is 700 micro coulombs, so just needs to be converted to coulombs

Thanks for your help. Annoyingly it still doesn't work as it becomes 1000 times too small.

From what I am hearing you need help converting. I always convert units for ease at the start. Just incase you don't know then to convert micro to normal it is times ten to the negative 6 (x10^-6)

Drawing capacitance/voltage graphs for capacitors in higher physics last year was the good life.


Posted from TSR Mobile

Yeahhhhhhh. Can't even remember fun like that. Good times...

Thanks for your help. Annoyingly it still doesn't work as it becomes 1000 times too small.

How's physics going for you so far ? I'm actually quite enjoying it partly because my teachers is sound as. About to finish unit 1 and probably have our nab soon after October break.


Posted from TSR Mobile

Nice. I did't take Physics this year (Bio, Chem, Maths). Pretty stupid idea considering Physics was my favourite subject last year!!!

I think you're having unit problems...

The best way to work is to *always* convert everything to SI units at the start, then do all your calculations, then convert back (if required).

The problem with working with a mixture of μF μA and whole volts is that sometimes it will appear to work correctly and sometimes you'll get an answer that is orders of magnitude out - which is exactly what's happened to you here

Thanks. I'm just confused because for example in part a. I've worked the answer out in microamps as this is the unit given in the question. The answer I got was 912 microamps but the answer is given as 0.91microamps.

Oh sorry - it's an error in your book... even had another thread about it a while ago... http://www.thestudentroom.co.uk/showthread.php?t=2021236