I Electrostatic potential a a point, the pluses and minuses

Tags:
1. Apr 25, 2016

Nikhil R

On calculating the electrostatic potential at a point due to charge q, by definition, it is the work done to bring a unit positive charge from infinity to that point. Trying to find it mathematically, it should be

∞→R ∫E.dr where E = k* q/r*r

On evaluating the integral, i find a minus sign in the answer which is -kq/R. I have a few doubts regarding this.

1.Is potential energy the work done by external force against the electrostatic force or the work done by the electrostatic force itself.

2. If there is a minus sign in the work done, it should be the case where the work done is by the system, that is without any external force, the body must move from initial to final position yielding that amount of energy. Which cannot be the case with a positive charge if the source charge at the centre is also positive.

Last edited: Apr 25, 2016
2. Apr 25, 2016

jbriggs444

Examine that result carefully. You have a definite integral in terms of R. You have expressed the result in terms of r.

3. Apr 25, 2016

Nikhil R

Thank you and apologies jbriggs444. I have made the correction, editing the post. Kindly consider the edited post.

4. Apr 25, 2016

jbriggs444

Be careful with your signs. This is a definite integral from ∞ to R. Should the result differ from a definite integral from R to ∞?

5. Apr 25, 2016

Nikhil R

On solving the integral

∞→R ∫E.dr where E = k* q/r*r , becomes,

= ∞→R ∫k*q/r*r dr

= k*q ∞→R∫1/r*r dr

= k*q [-1/r] upper limit - R lower limit - ∞

= k*q [-1/R - -1/∞]

= k*q [-1/R]

= -q*k/R

This i believe is the simplification of the integral. But in terms of the physics, there should not be a minus sign there. Am i missing something or is there any conceptual mistake in the way i take the integral.

6. Apr 25, 2016

jbriggs444

My apologies. Your work is sound. You have successfully calculated the work done by electrostatic repulsion as a unit charge is brought from infinity to a separation of R. Your original question:
is apt. The potential energy is the work done by the external force.

7. Apr 25, 2016

Nikhil R

I believe that i have calculated the work done by the applied force, because,on calculating the work done by the electrostatic repulsion which is towards infinity, the angle between displacement and the repulsive force must be 180 and E.r becomes - Er. The reason why i chose E.r as Er is because the applied force, which must be equal and opposite to the electrostatic repulsion takes place in the direction of the displacement and hence the angle between them is 0, making cosθ = 1.

Which is making me wonder why the work done by the applied force is obtained as negative which should come out as positive.

8. Apr 25, 2016

Nikhil R

Another question i have on this regard is, on applying definite integral to find work done, we integrate F.dx and apply lower and upper limits. Should we apply the dot product, before integration , that is -1 for θ = 180, 1 for θ = 0. Or will the limits applied and their values suffice. I have seen a famous lecture in which to find gravitational potential energy, on finding the work done in bringing a mass m from infinity to a point R , ∫F.dr is solved as ∫F*dr which actually should be a ∫-F*dr because applied force in that case is against gravitational attraction which must give cosθ as -1.

9. Apr 25, 2016

Nikhil R

I have been pondering on the above question for hours now. I must be missing some important basics here. Please kindly help me solve the confusion.

I believe that the sign of emerging from evaluation of any physical quantity should imply the physical meaning and vice versa. But this does not seem to work here.

10. Apr 25, 2016

Snigdh

I think that it is because you should take the sign of E in Edr as -be, after all, the direction of the displacement and the field is the opposite.

11. Apr 25, 2016

Nikhil R

Snigdh thank you for responding. I believe the sign of E.dr should be decided considering the mutual orientation of E and dr. Plus if they are parallel, and minus if they are anti parallel. In this case, the applied force is against the electrostatic force and the displacement is in the same direction as that of the applied force. In that case, applied force and displacement are in the same direction. and hence the angle will be zero, cosine will be 1.

12. Apr 25, 2016

SammyS

Staff Emeritus
The displacement is in a direction opposite to the direction of $\ \vec{dr}\$.

Therefore $\ \vec{ds}=-\vec{dr}\$.

13. Apr 25, 2016

Nikhil R

Thank you SammyS for responding to the question. I really appreciate your help.

Could you please help me understand how vector dr is defined. What i believed so far was that dr vector has an infinitely small magnitude and occurs in the direction of the motion that we choose. Therefore, the applied force in the following case is from point p towards the charge q1 and the displacement is in the same direction. To find the work done due to the applied force, i tried calculating the work done in moving the test charge or unit charge along the applied force for a length dr and then integrating it over infinity to R. My idea in all similar problems including that of gravitation is to consider the direction of dr in the direction of motion and then applying the limits initial to final. I must be missing something or may be wrong in some core basic, kindly help me figure it out.

q1°--------------------------------°P----------------------------------------------------------------∞

14. Apr 25, 2016

SammyS

Staff Emeritus
I assumed you were using standard dr, where r is the distance from the origin and the point charge is located at the origin.

When you write $\displaystyle \ E = k\frac{q}{r^2} \,,\$ dr is positive in the direction of increasing r, which in this case is away from the point charge.

15. Apr 25, 2016

Nikhil R

Thank you SammyS. But is dr a quantity that is defined by standard. i have always seen it as a quantity that we define for our problem with a liberty to chose the direction according to each problem. Kindly help me correct my mistake.

16. Apr 25, 2016

SammyS

Staff Emeritus
The issue is with what ever variable you are using to express the force, whether r or x or whatever.

17. Apr 25, 2016

Nikhil R

I intend to use the variable r and express force/ field at a point in terms of r.

18. Apr 25, 2016

SammyS

Staff Emeritus
Then as I said dr is positive in whatever direction r itself increases.

19. Apr 25, 2016

Nikhil R

So in this case , is it correct to say that applied force and -dr are in the same direction and hence, cosθ between them is 1.

20. Apr 25, 2016

Nikhil R

I mean the angle between F and [-dr].