Edexcel PHY 5 Revision - need help with any questions?

-6735 (01) - Physics.pdf275.7KB

Could you explain Q4a and 4bii) to me, i know the answers but not why....

Reply 22

ma3eeni

3

F=BIL
for F(z) = B(y)*I(z)*L
for F(y) = B(z)*I(y)*L
Since the force acting on each wire, is caused by the magnetic flux density of the other wire.
This is where I think you went wrong.

B=uI/2pi*r which could be rewritten to kI for simplicity, since u/2pi*r is a constant for both.

This gives F(z) = kI*2I*L
and F(y) = 2kI*I*L

Therefore F(z) : F(y) = 1 : 1

Reply 23

aelred89

A quick question, "Explain why a very small test charge must be used when measuring electric field strength" any ideas? I'm sure the answer is so simple but i just cant see it lol

Reply 24

Method

12

I think its something to do with, it doesn't produce a large field on its own which may affect the electrical field. :s-smilie:

Reply 25

OP

aelred89

A quick question, "Explain why a very small test charge must be used when measuring electric field strength" any ideas? I'm sure the answer is so simple but i just cant see it lol

I don't see why you have to use a test charge at all... you could use V/d for uniform electric field strength for electrical plates or kQ/r^2 for point charge.

I guess it's used to determine the direction of the field strength?

Reply 26

aelred89

A quick question, "Explain why a very small test charge must be used when measuring electric field strength" any ideas? I'm sure the answer is so simple but i just cant see it lol

So it has no effect on the field.

Reply 27

OP

-hybrid-

So it has no effect on the field.

You're right. :redface:

Reply 28

tziek

I think its something to do with, it doesn't produce a large field on its own which may affect the electrical field. :s-smilie:

Sorry didnt see someone already replied.

Reply 29

Eau

You're right. :redface:

I just hope exam questions will be that easy, havent revised much and I need an A.

Reply 30

OP

-hybrid-

I just hope exam questions will be that easy, havent revised much and I need an A.

Who doesn't? :wink:

Reply 31

I did a paper earlier and I found it so hard, some of the questions were really confusing. One question was about a drop of oil between to parallel plates connected to a power supply, eventually it asked about the voltage required for the drop to be stationary between the plates, checked the mark scheme and the answer involved using Vq/d=mg. My brain just froze.

Reply 32

Now thinking about it, are you suppose to equate the two as at this moment the forces on the drop are balanced?

Reply 33

OP

-hybrid-

Now thinking about it, are you suppose to equate the two as at this moment the forces on the drop are balanced?

Yes.

Reply 34

aelred89

Oh god my brain is going to explode :frown:

Reply 35

aelred89

Hi, another explainy question lol I know whats happening i just can't get it onto the paper,

"A capacitor is connected in series with a battery, a resistor and a micro-ammeter. Explain why a current flows initially in the rest of the circuit despite the fact that no charge is able to pass directly between the capacitor's plates."

I've got the ball resevoir analagy in my head so i understand that battery draws the charge from plate and transfers an equal charge to the other plate but where would i go from there?

Reply 36

Mohit_C

10

Well I would think that as it draws charge from one plate and transfers it equally to the other plate via the rest of the circuit, there already is a current as current is equal to the rate of flow of charge. The charge can't go directly pass the plates as there is an insulating material (a dielectric I think).

Reply 37

OP