Physics Concept.

[Physics Nerd]

I've been a lurker for a while, heard about the site from a friend.

Anyway, my physics teacher has given us a problem (not HW, just wants to see what we think it is tomorrow in class) and I can't figure it out. It's conceptual and he said we can come to him to see if we got it right but I don't see how it makes sense.

Here's the question:

Consider two charged spheres, one with charge +2C and the other with -2C. A proton is at the point halfway between the spheres. What is not zero?

a) the potential energy of the proton
b) the work to move the proton from infinity to that point
c) the force on the proton
d) all of the three above are zero.

I think it's a, b, and c but there's only one answer.

Reasoning for a):
There's an electric field created, right? Electric potential energy = (electric field)*(charge) so that has to be non-zero.

Reasoning for b):
Work is the negative of Electric Potential Energy, so if "a" is non-zero, "b" must be as well.

Reasoning for c):
The Electric Field exerts a force on the proton because the proton has a charge.

How can it be just one??? Please help

Help is greatly appreciated

 

[dude8604]

All are zero. The +2C and -2C charges have opposite forces on the proton.

a) The potential energy from one of the charges by itself is nonzero, but when you take both into account they cancel each other out. So the net potential energy is zero.

b) Since electric potential at infinity is usually defined as zero, and since the proton has no potential energy it has no electric potential. So the difference in electric potential is zero, so no work is done.

c) The NET force on the proton is zero, but there are two opposing forces acting on it. So it's possible this is a trick question since he didn't put net force.

d) Most likely is the answer.

edit: Btw, what level physics is this?

 

[NikolaeVarius]

Its D

 

[BassBomb]

Dude is right cept for a little

b) It is not because the point has no potential, but that one can travel along the equipotential surfaces from infinity. By definition of equipotential surface, traversing across them requires no work to be done.


All are zero, thus d) is true

 

[Rockinacoustic]

Shens on your name

 

[naldo]

lol, no wonder I barely passed physics, I figured it would be b since protons have a positive charge right? so the proton would be forced away from the positive sphere and attarcted to the negative sphere. and b has terms/concepts that I dont understand, so that would have to be it!

 

[Physics Nerd]

the way i see it, the +2 charge should "push" the proton towards the -2 charge and the -2 charge should "pull" the proton towards itself. what am i missing?

 

[Mo0o]

Originally posted by: Physics Nerd
the way i see it, the +2 charge should "push" the proton towards the -2 charge and the -2 charge should "pull" the proton towards itself. what am i missing?

Yeah thats kind of what I'm thinking since the proton is between teh sphere and not to either side of the pair...

 

[WildHorse]

``

 

[Physics Nerd]

Originally posted by: scott
foot bone connects to the

ANKLE bone

ankle bone connects to the

LEG bone

leg bone connects to the
HIP bone

and hokey pokey yourself around

that's all the physics you need

hate to break it to ya, but that sounds more like human physiology to me

 

[Fenixgoon]

someone correct me if im wrong, but shouldn't the +2 charge should be repelling the proton while the -2 charge attracts it, creating a net force towards the -2 charge? you would need 2 equal negative charges opposite each other for the net force to be 0.

apparently electrostatic potential energy, U, = kq1q2/r. the net potential is zero since the charges are equal and opposite - you have k*(+1)*(-2)/r + k*(+1)*(+2)/r. the work required to move to infinity would be equal to the potential energy... thus, also 0.

 

[Physics Nerd]

Originally posted by: Fenixgoon
someone correct me if im wrong, but shouldn't the +2 charge should be repelling the proton while the -2 charge attracts it, creating a net force towards the -2 charge? you would need 2 equal negative charges opposite each other for the net force to be 0.

apparently electrostatic potential energy, U, = kq1q2/r. the net potential is zero since the charges are equal and opposite since you have k*(+1)*(-2)/r + k*(+1)*(+2)/r. the work required to move to infinity would be equal to the potential energy... thus, also 0.

i think the equation uses the magnitudes of the charge, so it would be k*(+1)*(+2)/r + k*(+1)*(+2)/r. and the denominator is r^2?

 

[Fenixgoon]

Originally posted by: Physics Nerd

Originally posted by: Fenixgoon
someone correct me if im wrong, but shouldn't the +2 charge should be repelling the proton while the -2 charge attracts it, creating a net force towards the -2 charge? you would need 2 equal negative charges opposite each other for the net force to be 0.

apparently electrostatic potential energy, U, = kq1q2/r. the net potential is zero since the charges are equal and opposite since you have k*(+1)*(-2)/r + k*(+1)*(+2)/r. the work required to move to infinity would be equal to the potential energy... thus, also 0.

i think the equation uses the magnitudes of the charge, so it would be k*(+1)*(+2)/r + k*(+1)*(+2)/r. and the denominator is r^2?

you definitely need to use the charge values (positive/negative). a negative force indicates attraction while a positive indicates repulsion in electrostatics.

kq1q2/r^2 is for force. single R is (electrostatic potential) energy. F = dE/dr

 

[Mo0o]

Originally posted by: Fenixgoon

Originally posted by: Physics Nerd

Originally posted by: Fenixgoon
someone correct me if im wrong, but shouldn't the +2 charge should be repelling the proton while the -2 charge attracts it, creating a net force towards the -2 charge? you would need 2 equal negative charges opposite each other for the net force to be 0.

apparently electrostatic potential energy, U, = kq1q2/r. the net potential is zero since the charges are equal and opposite since you have k*(+1)*(-2)/r + k*(+1)*(+2)/r. the work required to move to infinity would be equal to the potential energy... thus, also 0.

i think the equation uses the magnitudes of the charge, so it would be k*(+1)*(+2)/r + k*(+1)*(+2)/r. and the denominator is r^2?

you definitely need to use the charge values (positive/negative). a negative force indicates attraction while a positive indicates repulsion in electrostatics.

kq1q2/r^2 is for force. single R is (electrostatic potential) energy. F = dE/dr

That's what I was thinking so not sure why the 1st 3 replies said D

 

[WildHorse]

Originally posted by: Physics Nerd

Originally posted by: scott
foot bone connects to the

ANKLE bone

ankle bone connects to the

LEG bone

leg bone connects to the
HIP bone

and hokey pokey yourself around

that's all the physics you need

hate to break it to ya, but that sounds more like human physiology to me

from my heart and from my hands
why don't (you) people understand????

weird!


Science

 

[dighn]

Originally posted by: Mo0o

Originally posted by: Fenixgoon

Originally posted by: Physics Nerd

Originally posted by: Fenixgoon
someone correct me if im wrong, but shouldn't the +2 charge should be repelling the proton while the -2 charge attracts it, creating a net force towards the -2 charge? you would need 2 equal negative charges opposite each other for the net force to be 0.

apparently electrostatic potential energy, U, = kq1q2/r. the net potential is zero since the charges are equal and opposite since you have k*(+1)*(-2)/r + k*(+1)*(+2)/r. the work required to move to infinity would be equal to the potential energy... thus, also 0.

i think the equation uses the magnitudes of the charge, so it would be k*(+1)*(+2)/r + k*(+1)*(+2)/r. and the denominator is r^2?

you definitely need to use the charge values (positive/negative). a negative force indicates attraction while a positive indicates repulsion in electrostatics.

kq1q2/r^2 is for force. single R is (electrostatic potential) energy. F = dE/dr

That's what I was thinking so not sure why the 1st 3 replies said D

easy mistake to make I guess, I made it too...

the answer is C

 

[BrownTown]

Well, I KNOW "C" isn't zero, so I'll go with that one.

 

[WildHorse]

Originally posted by: BrownTown
Well, I KNOW "C" isn't zero, so I'll go with that one.

well then maybe "A" his hiding that zero in her jacket

mmmm I don't know, they don't look real to me

 

[fawhfe]

Well I can't come up with any reason they all non-zero. If you look at how the problem is worded, it makes a fair bit of sense that the professor meant to write all 3 are non-zero since the question asks which are non-zero.

You reasoning seems sound other than on (a). Consider the case where both the "fixed" charges are +2P instead of +/-2P. Then there will be no electric field where the proton is, but the proton still has potential energy because if it were perturbed even slightly, it would go accelerating off perpendicular to the line between the two fixed charges. It's sort of like if you balanced a pyramid on it's tip. Just because the forces acting on it are balanced doesn't mean that it has no potential energy (since the pyramid could fall over, converting potential gravitational energy into kinetic energy).

 

[frostedflakes]

I think you guys are right, it's C. A is zero and therefore B is zero (because W=delta(U), and U at infinity is zero). The electric field, however, is not zero, and because F=qE, the force is non-zero as well.