The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Force on electric field with distance

Untitled.png

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? : Inverse.Proportion.Sm.Graph.gif

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?
Original post by You-know-who
Untitled.png

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? : Inverse.Proportion.Sm.Graph.gif

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.
(edited 9 years ago)
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! :smile:
Original post by You-know-who
Oh. Thank you so much! :smile:


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.
(edited 9 years ago)
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? :smile: 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. :biggrin:
Original post by You-know-who
I think you mean V/d when you say E/d? :smile: 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. :biggrin:
Yes of course - it was early morning! :tongue: :smile:

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!
(edited 9 years ago)
Original post by uberteknik
Yes of course - it was early morning! :tongue: :smile:

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. :tongue:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you