# Physics electrostatics

Watch
Announcements
#1

Can anyone tell me why the answer is C instead of D? Thanks!

and one more question: my teacher told me that when a charge of fixed charge density is put between two charges paralla plates (of different charge), THe Electric field is independent of the area of plates and the separation between plates.
but there is also an equation of E=V/d.

aren't they contradict with each other? can someone explain it to me???
thank you!!!!!!

Posted from TSR Mobile
0
6 years ago
#2
(Original post by Lamalam)

Can anyone tell me why the answer is C instead of D? Thanks!
Posted from TSR Mobile
Hi, what did you use for the circle questions?

Thanks!
0
#3
(Original post by Zenarthra)
Hi, what did you use for the circle questions?

Thanks!
what circle question??
0
6 years ago
#4
(Original post by Lamalam)
what circle question??
I think question 2. 0
#5
(Original post by Zenarthra)
I think question 2. Posted from TSR Mobile
0
6 years ago
#6
Thank you, but what is the 1.6 in the equation?
0
#7
(Original post by Zenarthra)
Thank you, but what is the 1.6 in the equation?
I sub 1.6 as the g because the question states that the loop is situated on moon. btw do you know the answer to my questions??
0
6 years ago
#8
(Original post by Lamalam)
I sub 1.6 as the g because the question states that the loop is situated on moon. btw do you know the answer to my questions??
Thank you for that!
I think, it is 2 points because you place a positive test charge on the line.
See here:

The one in red, the forces act in opposite direction and the black one they act in same direction.
But im not sure sorry. 0
6 years ago
#9
(Original post by Lamalam)

Can anyone tell me why the answer is C instead of D? Thanks!

and one more question: my teacher told me that when a charge of fixed charge density is put between two charges paralla plates (of different charge), THe Electric field is independent of the area of plates and the separation between plates.
but there is also an equation of E=V/d.

aren't they contradict with each other? can someone explain it to me???
thank you!!!!!!

Posted from TSR Mobile

You may have misheard your teacher.
The capacitance of a parallel plate capacitor is given by
C = ϵA/d to a good approximation, so that depends on the area of the plates.
The field between them is given by E = V/d as you say.
Those are the facts.
So the field shouldn't depend on the area of the plates so long as you do not go too near the edges of the plates where the field lines start to bend.
The field strength E does depend on the separation so long as V is constant. You need to check with your teacher exactly what he/she was actually referring to.
0
#10
(Original post by Stonebridge)
You may have misheard your teacher.
The capacitance of a parallel plate capacitor is given by
C = ϵA/d to a good approximation, so that depends on the area of the plates.
The field between them is given by E = V/d as you say.
Those are the facts.
So the field shouldn't depend on the area of the plates so long as you do not go too near the edges of the plates where the field lines start to bend.
The field strength E does depend on the separation so long as V is constant. You need to check with your teacher exactly what he/she was actually referring to.

This is stated in my note What is he trying to express?

Posted from TSR Mobile
0
#11
(Original post by Zenarthra)
Thank you for that!
I think, it is 2 points because you place a positive test charge on the line.
See here:

The one in red, the forces act in opposite direction and the black one they act in same direction.
But im not sure sorry. what are the first two sentences about?
and the third line: what is E referring to ?? the total electric field??

thankksss!!! I start to understnad a bit now!! 0
6 years ago
#12
(Original post by Lamalam)

This is stated in my note What is he trying to express?

Posted from TSR Mobile
Well the text says a constant charge density on the plates. Not a constant charge between the plates. Yes E depends on d if V is kept constant, as I said. But reducing d for example will increase the capacitance. Halving d doubles the capacitance according to the formula C = ϵA/d. If you double the capacitance with constant Q then you halve the pd from Q = CV
So you halve the distance d but the result is you also halve the pd.
So E doesn't change.
0
6 years ago
#13
(Original post by Lamalam)
what are the first two sentences about?
and the third line: what is E referring to ?? the total electric field??

thankksss!!! I start to understnad a bit now!! Yes if you look at the red writing, you can see that the force acting on the test positive charge will be opposite in direction to each other.
So maybe there is a point where Fa force due to point charge A on the positive test charge and Fb the force due to point charge B on the positive test charge are equal in magnitude but not direction.
E at a point = Resultant F on positive test charge at that point / charge of positive test charge q
As you can see positive test charge is constant q
So if forces at a point due to each point charge are equal the Electric field due to each point charge at that point are equal.
Force direction is dependent on the charges.
Electric field is vector quantity so at that point you would have E=0 if you add the electric fields together.

For the Black writing, if the positive test charge is between the 2 point charges, the positive test charge will experience a force due to point charge A and B.
The direction of the force is in the same direction.
E at a point = Resultant F on positive test charge at that point / charge of positive test charge q.
They are in the same direction so the resultant electric field at some point between them can never be 0.
hence the electric field at some point between them can not be = 0
But Fa could = Fb or just 2Fa or 2Fb, so that the resultant force = Fa + Fb.
Where the electric field at any point between the point charges = Fa + Fb / q
because q is constant. ! No problem!
0
#14
(Original post by Zenarthra)
Yes if you look at the red writing, you can see that the force acting on the test positive charge will be opposite in direction to each other.
So maybe there is a point where Fa force due to point charge A on the positive test charge and Fb the force due to point charge B on the positive test charge are equal in magnitude but not direction.
E at a point = Resultant F on positive test charge at that point / charge of positive test charge q
As you can see positive test charge is constant q
So if forces at a point due to each point charge are equal the Electric field due to each point charge at that point are equal.
Force direction is dependent on the charges.
Electric field is vector quantity so at that point you would have E=0 if you add the electric fields together.

For the Black writing, if the positive test charge is between the 2 point charges, the positive test charge will experience a force due to point charge A and B.
The direction of the force is in the same direction.
E at a point = Resultant F on positive test charge at that point / charge of positive test charge q.
They are in the same direction so the resultant electric field at some point between them can never be 0.
hence the electric field at some point between them can not be = 0
But Fa could = Fb or just 2Fa or 2Fb, so that the resultant force = Fa + Fb.
Where the electric field at any point between the point charges = Fa + Fb / q
because q is constant. ! No problem!
Ho lek ah nei thanks!

Posted from TSR Mobile
0
6 years ago
#15
(Original post by Lamalam)
Ho lek ah nei thanks!

Posted from TSR Mobile
wey, ho do je nei gum jang oh ah, dang hui ow doe mm hiu gum lek gah xD
Wey, ley doe how lek ah! 0
#16
(Original post by Stonebridge)
Well the text says a constant charge density on the plates. Not a constant charge between the plates. Yes E depends on d if V is kept constant, as I said. But reducing d for example will increase the capacitance. Halving d doubles the capacitance according to the formula C = ϵA/d. If you double the capacitance with constant Q then you halve the pd from Q = CV
So you halve the distance d but the result is you also halve the pd.
So E doesn't change.
I am still a bit confused can you explain more ? thank you
0
6 years ago
#17
(Original post by Lamalam)
I am still a bit confused can you explain more ? thank you
I'm not sure what you are stuck on.

If you halve d, the distance between the plates, and the result is that you also halve V, the pd between the plates, then the field strength E doesn't change because E = V/d
All you've done is get 0
#18
(Original post by Stonebridge)
I'm not sure what you are stuck on.

If you halve d, the distance between the plates, and the result is that you also halve V, the pd between the plates, then the field strength E doesn't change because E = V/d
All you've done is get I thought there is only electric field strength between the the parallal plates, there is also electric field strength on the plate?!
0
6 years ago
#19
(Original post by Lamalam)
I thought there is only electric field strength between the the parallal plates, there is also electric field strength on the plate?!

Are you sure you understand what electric field strength is?
0
#20
(Original post by Stonebridge)
Are you sure you understand what electric field strength is?

A region of space in which a charged object experience a force?
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Should there be a new university admissions system that ditches predicted grades?

No, I think predicted grades should still be used to make offers (675)
33.62%
Yes, I like the idea of applying to uni after I received my grades (PQA) (856)
42.63%
Yes, I like the idea of receiving offers only after I receive my grades (PQO) (386)
19.22%
I think there is a better option than the ones suggested (let us know in the thread!) (91)
4.53%