Electric potential energy due to a point charge

[Ashu2912]

Hey friends, I am stuck up in the derivation of the electric potential energy due to a charge 'Q' at the origin on test charge 'q' at a point p (position vector 'r_p'). The derivation is shown below, am just struggling with a minus sign...

Electric Potential Energy at P
= Work done to move 'q' from infinity to P quasi-statically by external agent

=\int kQq/r² -r^\wedge.-dr

(Since External Force = - Coulumbic Force and displacement vector = -dr)

=kQq [-1/r]_\infty^rp

=-kQq/r

I know that when Qq is +ve, work done should be positive, but the sign is changing due to the integral.

NOTE : all vectors shown in bold faces, r^ is unit vector along r and dr is infinitesimal displacement in direction of r...

 

[rcgldr]

Potential energy is the negative of the work done by the force of a field. Could the be the issue you're having?

 

[Ashu2912]

Ya, that's true but here I am considering the work done by an external agent, which provides force just enough to counter-balance the force of the electric field. Therefore, this external agent must do positive work in case of repulsive forces (Qq is positive) as the potential energy of the charge 'q' is increased on it's action...

 

[rcgldr]

Since Qq is positive, the external force must be doing negative work (and the field doing positive work) when allowing the charges to move apart.

 

[Ashu2912]

Ya, I got your point. However, my textbook defines electrostatic potential energy as
"Potential energy of a charge q at a point in the presence of field due to any charge configuration is the work done by the external force (equal and opposite to the electric force) in bringing a charge q from infinity to that point".
Now, if we go to see
Work done by external force in moving charge from \infty to the point
= W_\inftyP
= U_initial - U_final
= -U_final

or Electrostatic potential U_final = -W_\inftyP (by external force), which is a contradiction to the definition.

My guess if that the work done by the external force should be positive, in case of let's say Q,q > 0, but it is coming to be negative, equal to -kQq/r. Even though the infinitesimal work is coming to be positive, the net work is coming negative. Don't know why. Please help!

 

[Doc Al]

Looks to me like your original derivation is off a bit. Note that 'dr' points in the direction of increasing r, so the work done against the field is:
dW = -qEdr (dot product, of course)

For a point charge Q, the potential energy at r is the work done against the field from infinity to point r:
U(r) \equiv -\int_{\infty}^r q\vec{E} \cdot d\vec{r}

U(r) \equiv -\int_{\infty}^r \frac{kQq}{r^2} dr = \frac{kQq}{r}

 

[Ashu2912]

However, the infinitesimal work done should be
dw = F_ext.ds (ds is in the direction of \infty to r_p)
Since ds = -dr and F_ext is -F_electric
dw = F_electric.dr, which is positive, assuming Q,q>0

 

[Ashu2912]

I'm once again showing my complete derivation, with the diagram attached. The diagram is made in MS Paint, so...my apologies. All vectors are shown in bold face.
Consider a charge Q>0 at origin. We bring another charge q>0 from infinity to P, with position vector p. At each instant we apply an external force, equal and opposite to the Coulumbic force, so that the charge q comes from infinity to P quasi-statically, with constant velocity. Now,
Work done in moving the charge q from infinity to P
= w_\inftyP

=\intF_external.ds from \infty to p

=\int-F_electric.-dr (as ds = -dr. Since r is the general position vector at a point directed from the origin, dr too, points away from the origin and is parallel to r, and joins r to r + dr. ds is used in the main equation as we have to integrate along the path followed).

=\intkQq/r² dr (F_electric and dr are in the same direction, so their dot product is Fdrcos0 = Fdr)

=-kQq/p

I have one more confusion. We know that \DeltaU = U_final - U_initial = -Work done by a force. Using this, we get U_final = kQq/p, since U_initial is taken as 0.

However my textbook says that work done by external force to move charge from A to B =
U_B - U_A.

Please solve my big confusion. My exam's on Monday. Pls. help as fast as you can...

 

[Doc Al]

I'm once again showing my complete derivation, with the diagram attached. The diagram is made in MS Paint, so...my apologies. All vectors are shown in bold face.
Consider a charge Q>0 at origin. We bring another charge q>0 from infinity to P, with position vector p. At each instant we apply an external force, equal and opposite to the Coulumbic force, so that the charge q comes from infinity to P quasi-statically, with constant velocity. Now,
Work done in moving the charge q from infinity to P
= w_\inftyP

=\intF_external.ds from \infty to p

=\int-F_electric.-dr (as ds = -dr. Since r is the general position vector at a point directed from the origin, dr too, points away from the origin and is parallel to r, and joins r to r + dr. ds is used in the main equation as we have to integrate along the path followed).

=\intkQq/r² dr (F_electric and dr are in the same direction, so their dot product is Fdrcos0 = Fdr)

=-kQq/p

Too many minus signs.

The work done would be:
\int_\infty^r \vec{F_{ext}}\cdot d\vec{s}

Since s points toward the charge, the force F_ext is positive, not negative.

\int_\infty^r \frac{kQq}{r^2} ds = -\int_\infty^r \frac{kQq}{r^2} dr

I have one more confusion. We know that \DeltaU = U_final - U_initial = -Work done by a force. Using this, we get U_final = kQq/p, since U_initial is taken as 0.

However my textbook says that work done by external force to move charge from A to B =
U_B - U_A.

I guess I don't understand the question. The work required to move a charge quasi-statically from A to B will equal ΔU = U_B - U_A.

 

[Ashu2912]

Work done =
\int_\infty^r \frac{kQq}{r^2} ds = -\int_\infty^r \frac{kQq}{r^2} dr

In the above equation, F_ext and ds are in the same direction. The above integral shouldn't therefore contain |F_ext||ds|cos0
=kQq/r² X dr, as ds = -dr => |ds| = |dr|. Therefore, shouldn't the final expression contain a minus sign? I understand that since F_ext and ds are in the same direction, thus the work done by the external force should be positive. However, the infinitesimal work is coming to be positive, but the total work is coming to be negative, just because of integration...

In my second question, I meant that \DeltaU = -W, right? So, U_i - U_f = W_{i to f}, as per the work - PE relation. But in my book, it is given that U_f - U_i = W. So, which of them is correct?

 

[Doc Al]

Work done =
\int_\infty^r \frac{kQq}{r^2} ds = -\int_\infty^r \frac{kQq}{r^2} dr

In the above equation, F_ext and ds are in the same direction. The above integral shouldn't therefore contain |F_ext||ds|cos0
=kQq/r² X dr, as ds = -dr => |ds| = |dr|. Therefore, shouldn't the final expression contain a minus sign?

I'm not sure what you mean by 'final expression', but the last expression in the quoted segment does have a minus sign.

I understand that since F_ext and ds are in the same direction, thus the work done by the external force should be positive.

Right.

However, the infinitesimal work is coming to be positive, but the total work is coming to be negative, just because of integration...

No. How can the sum of positives end up negative?

In my second question, I meant that \DeltaU = -W, right?

That depends on what force is doing the work. In the definition of potential energy that we are discussing above, it's an external force doing the work. Thus ΔU = W, not -W.

 

[Ashu2912]

The last expression i meant was:
dW = kQq/r²ds
In this, kQq/r²ds is a scalar expression, a dot product. Since ds=-dr (please check my diagram again) so |ds| = |dr|. Thus, dW is coming to be +ve.
How can the sum of positives end up negative?
Thats what I am thinking.
Thanks, I got the second part. Thanks a lot!

 

[Doc Al]

The last expression i meant was:
dW = kQq/r²ds

OK.

In this, kQq/r²ds is a scalar expression, a dot product.

Right. And F and ds point in the same direction, so Fds is positive.

Since ds=-dr (please check my diagram again) so |ds| = |dr|. Thus, dW is coming to be +ve.

This is confusing. Fds is positive. Now you want to change your variable of integration from s to r, so use ds = -dr (not ds = dr).

 

[Ashu2912]

Ya, but in terms of magnitude, ds = dr, since ds = -dr, right? Thanks so much for your time...

 

[Doc Al]

Ya, but in terms of magnitude, ds = dr, since ds = -dr, right?

Well, sure. The magnitude of ds and dr are equal. But you cannot ignore the sign!

 

[Ashu2912]

I have finished with my exam, but still have one doubt in my mind...
When we have a limit of integration like [-1/r] from \infty to let's say any positive number, we get a negative number, even though the infinitesimal quantity is positive, since the direction of integration also matters. So do you mean that we will have to change the sign of the integration variable so that the work done comes to be positive??
However, if this is the case, then why do we get a negative sign even when we integrate from -\infty to a positive number?
I'm oinfused totally in this matter. Please help!

 

[Doc Al]

I have finished with my exam, but still have one doubt in my mind...
When we have a limit of integration like [-1/r] from \infty to let's say any positive number, we get a negative number, even though the infinitesimal quantity is positive, since the direction of integration also matters.

I assume you mean to integrate -1/r^2, not -1/r. In any case, why do you think the integral from ∞ to some positive number would be negative?

 

[dreamLord]

Bumping this because I have exactly the same doubt, and reading through this thread didn't help me clear it :/

We know that F(ext) = -qE, where E is the field due to a positive charge along 'r' vector. Also, our displacement in moving the test charge near the original charge is along 'r' vector, but in the opposite direction, hence '-dr'.

Now dW = F(ext).ds = (-qE) * (-dr), integration of which gives me a negative quantity. Where have I made a mistake of a minus sign?

 

[Ashu2912]

Hey dreamlord, I think I have understood this now!
Fext(vector) =- qE(vector) like you said. However, while integrating you have to consider dr(vector) only, the integration limits being from 0 to infinity!

 

[Doc Al]

We know that F(ext) = -qE, where E is the field due to a positive charge along 'r' vector. Also, our displacement in moving the test charge near the original charge is along 'r' vector, but in the opposite direction, hence '-dr'.

Now dW = F(ext).ds = (-qE) * (-dr), integration of which gives me a negative quantity. Where have I made a mistake of a minus sign?

The sign of dr is taken care of by your limits of integration. You move in the '-dr' direction by integrating from ∞ to r.

 

[Ashu2912]

Hey dreamlord, I think I have understood this now!
Fext(vector) =- qE(vector) like you said. However, while integrating you have to consider dr(vector) only, the integration limits being from 0 to infinity!

Sorry I mean infinity to zero! And what doc al said is absolutely correct!