The Student Room Group

Electric potential

Hey,

If electric potential is a scalar quantity, then why is it negative for negative charges and positive for positive charges? Is this just simply due to the convention that its defined in terms of positive charges?

Thanks!

@uberteknik
Original post by Nikhilm
Hey,

If electric potential is a scalar quantity, then why is it negative for negative charges and positive for positive charges? Is this just simply due to the convention that its defined in terms of positive charges?

Thanks!

@uberteknik


Exactly.

Charge is a fundamental property of both Electrons and Protons. It is the description of a force acting over a distance where like charges repel and opposite charges attract.

The labelling of charge (attributed to Benjamin Franklin) and the direction of so called conventional-current flow is a purely arbitrary decision made long before the discovery of the atom and it's composition of a nucleus with accompanying oppositely charged electrons.

We now know that actual electron current (mobile charge carriers in a conductor) is always from the electrode with a glut of -ve charge (cathode) to the electrode with a paucity of -ve charge (anode).

The Franklin conventional-current describes current flowing in the opposite direction to actual flow. This convention stuck because by the time electrons were discovered, it's use was far too widespread (and hence difficult) to change.

An analogy would be, say, the convention of traffic lights (red is stop and green is go) or the north and south pole, east and west, left and right, up and down etc. etc. All purely arbitrary conventions.
(edited 7 years ago)
Original post by uberteknik
Exactly.

Charge is a fundamental property of both Electrons and Protons. It is the description of a force acting over a distance where like charges repel and opposite charges attract.

The labelling of charge (attributed to Benjamin Franklin) and the direction of so called conventional-current flow is a purely arbitrary decision made long before the discovery of the atom and it's composition of a nucleus with accompanying oppositely charged electrons.

We now know that actual electron current (mobile charge carriers in a conductor) is always from the electrode with a glut of -ve charge (cathode) to the electrode with a paucity of -ve charge (anode).

The Franklin conventional-current describes current flowing in the opposite direction to actual flow. This convention stuck because by the time electrons were discovered, it's use was far too widespread (and hence difficult) to change.

An analogy would be, say, the convention of traffic lights (red is stop and green is go) or the north and south pole, east and west, left and right, up and down etc. etc. All purely arbitrary conventions.


Thank you!
Original post by uberteknik
Exactly.

Quick question .... for F = Bqv, is v technically the speed (not the velocity) due to F being the magnitude of the force?

@uberteknik
[QUOTE="Nikhilm;63806981"]
Original post by uberteknik
Exactly.

Quick question .... for F = Bqv, is v technically the speed (not the velocity) due to F being the magnitude of the force?

@uberteknik


Strictly speaking, both the magnetic field (B) and velocity (v) are vectors since the force (vector) exerted on a charged particle is dependent on the relationship between its' velocity (vector) moving through the magnetic flux (vector).

The full definition is therefore:

F=B x qv\vec{F} = \vec{B}\ \mathrm x \ q \vec{v}

The truncated definition F = Bqv requires the assumption that velocity is always perpendicular to the magnetic field.
(edited 7 years ago)
[QUOTE="uberteknik;63811295"]
Original post by Nikhilm


Strictly speaking, both the magnetic field (B) and velocity (v) are vectors since the force (vector) exerted on a charged particle is dependent on the relationship between its' velocity (vector) moving through the magnetic flux (vector).

The full definition is therefore:

F=Bqv\vec{F} = \vec{B}q\vec{v}

The truncated definition F = Bqv requires the assumption that velocity is always perpendicular to the magnetic field.



Ah I see, a-level really dumbs this section down lol

One more q if you don't mind: regarding the 'magnet falling vertically through a coil' type --> since the speed of the magnet at the bottom of the coil is greater than its speed at the top, wouldn't this mean that the induced emf at the bottom of the coil is also greater than the induced emf at the top, and therefore that the induced current is greater at bottom that at top and so the ammeter would flick more?

Here's a link to a picture if that didn't make sense: http://moodle2.rockyview.ab.ca/pluginfile.php/72891/mod_book/chapter/31819/physics_30/images/m4/p30_m4_124_l.jpg

@uberteknik
[QUOTE="Nikhilm;63823435"]
Original post by uberteknik



Ah I see, a-level really dumbs this section down lol

One more q if you don't mind: regarding the 'magnet falling vertically through a coil' type --> since the speed of the magnet at the bottom of the coil is greater than its speed at the top, wouldn't this mean that the induced emf at the bottom of the coil is also greater than the induced emf at the top, and therefore that the induced current is greater at bottom that at top and so the ammeter would flick more?

Here's a link to a picture if that didn't make sense: http://moodle2.rockyview.ab.ca/pluginfile.php/72891/mod_book/chapter/31819/physics_30/images/m4/p30_m4_124_l.jpg

@uberteknik


Hey, would really appreciate some help on this when you get some time haha @uberteknik
Original post by uberteknik


The full definition is therefore:

F=Bqv\vec{F} = \vec{B}q\vec{v}


It should be F:=q(v×B)=B×qv\vec{F}:=q(\vec{v} \times \vec{B})= -\vec{B}\times q\vec{v}.
Original post by morgan8002
It should be F:=q(v×B)=B×qv\vec{F}:=q(\vec{v} \times \vec{B})= -\vec{B}\times q\vec{v}.
I can't grumble if you want to give it the full Lorentz cross-product treatment but it's a tad OTT for A-level physics.
(edited 7 years ago)
Original post by uberteknik
I can't grumble if you want to give it the full Lorentz cross-product treatment but it's a tad OTT for A-level physics.


But you've already given the cross-product expression. :confused:

You put the vectors in non-standard order though - that's what he's pointing out.
Original post by atsruser
But you've already given the cross-product expression. :confused:

You put the vectors in non-standard order though - that's what he's pointing out.
I'm pointing out that I doubt many A-level students will know what a cross product is let alone knowing how to apply it.
Original post by uberteknik
I'm pointing out that I doubt many A-level students will know what a cross product is let alone knowing how to apply it.


Most won't but a non-negligible portion will. Anyone who's studied the relevant modules in further maths or learnt it separately will know what it is and how to apply it. When I was doing A-level I found the cross product to be the easiest way to remember the direction of the magnetic force.
Original post by morgan8002
Most won't but a non-negligible portion will. Anyone who's studied the relevant modules in further maths or learnt it separately will know what it is and how to apply it. When I was doing A-level I found the cross product to be the easiest way to remember the direction of the magnetic force.


How would you go about doing that?
Original post by Nikhilm
How would you go about doing that?


If you don't know the cross product then don't worry.

If you do then you just need to work out the cross product forms of the A-level equations and remember them. eg. F=q(B×v), F=IL×B\vec{F} = q(\vec{B}\times\vec{v}),\ \vec{F} = I\vec{L}\times \vec{B}, where L is in the same direction as current.

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