# Force on electric field with distance

Watch
Announcements
#1

I'm stuck at the last part of this question which asks us to sketch the graph. According to F = Vq/x , shouldn't F be inversely proportional to x? So shouldn't it give a graph similar to this? :

However, the answer states that the graph should be a straight horizontal line with the value of F marked. :/ How does F remain constant if distance along the plates is varied?
0
5 years ago
#2
(Original post by You-know-who)

I'm stuck at the last part of this question which asks us to sketch the graph. According to F = Vq/x , shouldn't F be inversely proportional to x? So shouldn't it give a graph similar to this? :

However, the answer states that the graph should be a straight horizontal line with the value of F marked. :/ How does F remain constant if distance along the plates is varied?
Be careful, the part where it says 'placed just above the bottom plate' together with the last part of the question is a bit of tricky wording meant to catch out the unwary!

The electric field between the plates is uniform and the distance relates to the separation between the plates and NOT the charged particle distance from the plate.

i.e. the charge experiences the same force when placed anywhere in the uniform electric field.
1
#3
(Original post by uberteknik)
Be careful, the part where it says 'placed just above the bottom plate' together with the last part of the question is a bit of tricky wording meant to catch out the unwary!

The electric field between the plates is uniform and the distance relates to the separation between the plates and NOT the charged particle distance from the plate.

i.e. the charge experiences the same force when placed anywhere in the uniform electric field.
Oh. Thank you so much!
0
5 years ago
#4
(Original post by You-know-who)
Oh. Thank you so much!
Think of it this way:

The electric field varies from one plate to the other at a constant rate of

E/d = 3000 / 0.025 = 120kV / m

If the charge is placed at 10mm distance from the 0V plate, then the potential at that position will be:

(10 / 25 ) x 3000 = 1200V

which means the force experienced by the charge will be:

F = Eq/d = 1200V x 1.6x10-19C / 0.010m = 1.92x10-14N

i.e. E / d is constant throughout the field because the electric potential at that position (between two parallel plates) is also a linear function of distance.
0
#5
(Original post by uberteknik)
Think of it this way:

The electric field varies from one plate to the other at a constant rate of

E/d = 3000 / 0.025 = 120kV / m

If the charge is placed at 10mm distance from the 0V plate, then the potential at that position will be:

(10 / 25 ) x 3000 = 1200V

which means the force experienced by the charge will be:

F = Eq/d = 1200V x 1.6x10-19C / 0.010m = 1.92x10-14N

i.e. E / d is constant throughout the field because the electric potential at that position (between two parallel plates) is also a linear function of distance.
I think you mean V/d when you say E/d? I use E for electric field strength so that the equation becomes V = Ed. But anyways.

So basically, the electric field strength at all points is the same since this is a uniform magnetic field. However, the potential difference varies, such that the ratio V/d is always constant. This is also obvious since V/d = E, and E is always constant. Since F = Eq, F = Vq/d. So, despite changes in d, E still remains constant. Since q is the same, F remains constant too.

Thank you for clearing this up again.
0
5 years ago
#6
(Original post by You-know-who)
I think you mean V/d when you say E/d? I use E for electric field strength so that the equation becomes V = Ed. But anyways.

So basically, the electric field strength at all points is the same since this is a uniform magnetic field. However, the potential difference varies, such that the ratio V/d is always constant. This is also obvious since V/d = E, and E is always constant. Since F = Eq, F = Vq/d. So, despite changes in d, E still remains constant. Since q is the same, F remains constant too.

Thank you for clearing this up again.
Yes of course - it was early morning!

PS Do you really mean magnetic field? The moving charge, accelerated by the electric field, will set up it's own magnetic field but the question does not reference any magnetic property (the plates hold static charge)......and we're digressing now!
0
#7
(Original post by uberteknik)
Yes of course - it was early morning!

PS Do you really mean magnetic field? The moving charge, accelerated by the electric field, will set up it's own magnetic field but the question does not reference any magnetic property (the plates hold static charge)......and we're digressing now!
Hahahahahahaha, oops, no, I meant electric field, sorry.
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

#### Current uni students - are you thinking of dropping out of university?

Yes, I'm seriously considering dropping out (185)
14.2%
I'm not sure (59)
4.53%
No, I'm going to stick it out for now (381)
29.24%
I have already dropped out (37)
2.84%
I'm not a current university student (641)
49.19%