Electrostatics question (Work and Potential)

[Irishdoug]

Homework Statement

Please see picture as diagram required for answer. I am looking to see if my answers are correct as I am unsure if they are. Thankyou for your help.

Relevant Equations

The E-field in the x-direction is zero for the charge at y due to symmetry , and in the y-direction is negative.

W = F*scos$\theta$

a) From X -Y. The work done on the positive charge is negative as the displacement is in the negative y-direction i.e. It is a positive charge moving in parallel to a negative E-field: W= F*(-s) = (+)(-) = -

b) Y-Z. The work done is 0. The E-field in the x-direction is 0 as they cancel due to symmetry. $E_{y}$ is also 0 as the charge is moving perpendicular to the Electric field in the y-direction. That is, Fscos$\theta$ = Fscos($\frac{pi}{2})$ = 0. As there is no E-field there is no force to work against and the work done is 0.

c) E-potential for Y is greater than Z.

Y is equi-distant from both charges. As such, $Vy$=$\frac{2kq}{r}$. The potential for Z in the x-direction --> $Vz_{x}$ = $\frac{kq}{r}$ cos$\theta$.

cos$\theta$ = x/r . So $Vz_{x}$ = $\frac{kqx}{r^{2}}$

Likewise $Vz_{y}$ = $\frac{kqy}{r^{2}}$

As such: $Vz_{x}$ + $Vz_{y}$ < $Vy$ due to $r^{2}$ terms in the $Vz$ equation

d) It is positive. The new charge has an E-field twice as large as the current E-field. As such, we are now moving a positive charge in a positive E-field (that is moving a positive charge in the negative y-direction in an E-field pointing in the positive y-direction). Therefore we lose energy and do positive work.

 

[Master1022]

a) From X -Y. The work done on the positive charge is negative as the displacement is in the negative y-direction i.e. It is a positive charge moving in parallel to a negative E-field: W= F*(-s) = (+)(-) = -

I don't think is correct. So work done is given by: $Work Done = \vec F \cdot \vec x$ where $x$ is the displacement. We can think about what way the electric field is acting at that point (which I believe is downwards in the -ve y direction) - that makes sense from intuition as we would expect negative charges to attract a positive charge.

Now that we have found the direction of the force acting on the particle, we can evaluate our dot product by thinking whether the particle moves in the direction of the force (+ve work) or not (-ve work)

 

[Irishdoug]

Hi,

Thankyou for your response. So you are saying that the work done is positive as the charge is moved in the same direction as the force acting on it, ie both in the negative y-direction?

As such, the answer to d) will then be negative work as the charge is moved opposite to the direction of the force (which is now in the +ve y-direction)?

 

[Master1022]

Hi,

Thankyou for your response. So you are saying that the work done is positive as the charge is moved in the same direction as the force acting on it, ie both in the negative y-direction?

As such, the answer to d) will then be negative work as the charge is moved opposite to the direction of the force (which is now in the +ve y-direction)?

Yes, I believe the work is positive for part (a).

Yes, I agree with you for part (d).

Also, just something to think about for part (b) (I have not made a formal post about this as I need to still convince myself of the answer): the E-field in the x-direction only (necessarily) cancels out when $q_0$ is halfway between the -Q charges. If we were to move $q_0$ to the right slightly, would the horizontal components of the electric force still cancel each other out? (Personally, I am still thinking about this)

 

[Irishdoug]

Yes, I believe the work is positive for part (a).

Yes, I agree with you for part (d).

Also, just something to think about for part (b) (I have not made a formal post about this as I need to still convince myself of the answer): the E-field in the x-direction only (necessarily) cancels out when $q_0$ is halfway between the -Q charges. If we were to move $q_0$ to the right slightly, would the horizontal components of the electric force still cancel each other out? (Personally, I am still thinking about this)

Ok I think you may be correct.

So, the force from the -Q on the LHS diminishes by half at z. As such, the force from the -Q on the RHS is the "main" force we are acting against as we move to the right from the midpoint.

The E-field points towards the negative x-direction (the midpoint).

Thus we are moving "against" the force (as we are moving in the positive x-direction) and as such we have negative work.

Does that seem correct to you?

 

[Cutter Ketch]

From X -Y. The work done on the positive charge is negative

If the forces are in the direction of the movement, the system is doing work on the test charge. Positive is attracted to negative, so the test charge will move from X to Y all by itself. You could attach a rope to it and do some work with it.

b) Y-Z. The work done is 0. The E-field in the x-direction is 0 as they cancel due to symmetry. EyEyE_{y} is also 0 as the charge is moving perpendicular to the Electric field in the y-direction.

That’s all incorrect. Say you were in the middle of the two positive charges. At that location the forces symmetrically cancel. however, moving from there toward one charge or the other the net force does not cancel. With the 1/r^2 force the test charge will reduce it’s energy by moving toward either charge. Think of it like two attractive wells. The center is a saddle point, but the wells get deeper in either direction. Similarly, out at Y the x components cancel, but it is an unstable point. The well gets deeper moving closer to either charge.

c) E-potential for Y is greater than Z.

This question has an error due to the ”absolute value”. However, ignoring that, I agree with you.

Since we are free to choose 0 potential anywhere we like, the absolute value in the question makes it possible to get either answer.

Therefore we lose energy and do positive work.

Your answer to the question is wrong, but your statement is correct. Neither question A nor question D asks if YOU are doing work on the test charge. They ask if the charges are doing work on the test charge. If YOU have to do work on the test charge to move it, then the SYSTEM is doing negative work. It is opposing the motion.

 

[Irishdoug]

If the forces are in the direction of the movement, the system is doing work on the test charge. Positive is attracted to negative, so the test charge will move from X to Y all by itself. You could attach a rope to it and do some work with it.

Ok I am a little confused! If you could bear with me please.

Am I correct in saying the following; The E-field is in the -ve y-direction. The charge moves in the negative y-direction. Therefore there is positive work being done by the Electric force on the $q_{0}$.

The reason I said negative initially is I read this online:

"Thus, if we move a positive charge in the direction of the electric field then we do negative work (i.e., we gain energy). Likewise, if we move a positive charge in the opposite direction to the electric field then we do positive work (i.e., we lose energy). "

But what I gather from the rest of your answer is that we are looking at this from the perspective of the force moving the charge, as opposed to "me" moving it.

http://farside.ph.utexas.edu/teaching/302l/lectures/node32.html

Putting maths around this $W = Fscos\theta$ The Force is negative (as have a negative charge), displacement (with respect to the E-field) is positive as it points down and $q_{0}$ moves down, and $cos\theta$ = $cos\pi$ = -1.

So we have ##W = (-)(+)(-) = +W

Is this correct?

That’s all incorrect. Say you were in the middle of the two positive charges. At that location the forces symmetrically cancel. however, moving from there toward one charge or the other the net force does not cancel. With the 1/r^2 force the test charge will reduce it’s energy by moving toward either charge. Think of it like two attractive wells. The center is a saddle point, but the wells get deeper in either direction. Similarly, out at Y the x components cancel, but it is an unstable point. The well gets deeper moving closer to either charge.

So in this case, the electric force is doing negative work as the E-Field is in the -ve x-direction, but $q_{0}$ moves in the +ve x-direction.

This question has an error due to the ”absolute value”. However, ignoring that, I agree with you.

Since we are free to choose 0 potential anywhere we like, the absolute value in the question makes it possible to get either answer.

Ok good. Was my method correct in determining the answer, or did I just get lucky (provided the midpoint is at Y which is what I assumed)?

Your answer to the question is wrong, but your statement is correct. Neither question A nor question D asks if YOU are doing work on the test charge. They ask if the charges are doing work on the test charge. If YOU have to do work on the test charge to move it, then the SYSTEM is doing negative work. It is opposing the motion.

The work done by the electric force for part d) is negative. The E-field is the +ve y-direction however the charge moves in the -ve y-direction.

 

[Cutter Ketch]

Am I correct in saying the following; The E-field is in the -ve y-direction. The charge moves in the negative y-direction. Therefore there is positive work being done by the Electric force on the q0

Yes, that’s it. If a cart rolls down a hill gravity is doing positive work on the cart. You don’t have to do it. Gravity is doing it. You could hook the cart up to a mill and grind some corn. If you push the cart back up the hill you have to do some work. Your positive work against the negative work being done by gravity. It balances out, see? You are storing your energy in gravitational potential energy where what we mean by “potential” is the potential to do future work.

So in this case, the electric force is doing negative work as the E-Field is in the -ve x-direction, but q0q0q_{0} moves in the +ve x-direction.

No, I’m afraid not. Z is considerably closer to the right hand charge than Y was. If you calculate the 1/r^2 forces for the two charges you will find Z is deeper in the well. Moving from Y to Z is still down hill. Another way to look at it, at Y you have field lines pointing toward each charge. The x components cancel. But when you move away from the middle the x components no longer cancel. The nearer charge becomes stronger. The net field lines point toward the nearer charge. That means the net field lines in the middle are pointing down, but away from the middle they lean toward the nearer charge. Moving from Y to Z is moving with a component of the field, not against it. The field does work. Another way to look at it: which way is down hill? If you let go of your test charge, would it stay where it is, would it move away, or would it fly to one of the charges depending on which way it slightly perturbed from the exact middle?

Was my method correct in determining the answer, or did I just get lucky (provided the midpoint is at Y which is what I assumed)?

Well, I think you were on the right track, but you had some mistakes. For example you wrote:

 

[Cutter Ketch]

Oops. Wrong button. As I was saying ...

You wrote:

Y is equi-distant from both charges. As such, $Vy=\frac{2kq}{r}$

Which isn’t quite right. That is the potential you would have if you were r away from a single charge of 2q. As you approach these separated charges components of the forces are pulling in opposite directions, so you know you know you won’t get that full negative potential. Anyhow, I think you had the right idea

 

[Irishdoug]

Ok I get it now thanks for your help.