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I Finding work done using definite integrals

  1. Apr 25, 2016 #1
    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 in deciding the sign of the final value.
    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.
    The same confusion exists in case of finding out the electrostatic potential due to a positive charge.
    Kindly help me understand this.
     
  2. jcsd
  3. Apr 25, 2016 #2
    Look at that problem from mathematical point of view.
    Let's see some more complicated example, when you are integrating some vector field along a line that is not straight. [itex]\int \vec{E} d\vec{l}[/itex] . Now you can't integrate then do a scalar product, because you are integrating a component of that vector field along your line of integration, and in every point of space, that component can have different value. So always do dot product before the integration because dot product is the function that you are integrating. Hope it helped.
     
  4. Apr 25, 2016 #3
    Thank you avalanche. I really understand the point. It definitely means that we should apply the dot product, make sure the cos θ is substituted andthen evaluate the integral. That is where my doubt arises.

    Using this idea, i try to find out the electrostatic potential at a point R distance from a point charge +q. Which by definition is the work done in bringing a unit positive charge from infinity to the point R.

    ∞→R ∫E.dr where E = k* q/r*r , becomes, E = gives the force per unit positive charge.

    = ∞→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 result seems illogical because both the applied force and the displacement are in the same direction and the potential at that point is expected to be positive.
     
  5. Apr 25, 2016 #4
    I get the same sign inverted answer when i try doing the same problem for Gravitational Potential Energy. The work done in bringing a point mass from infinity to a pint R . In this case The applied force is against the attractive gravitational force. F applied is outward and dr is inward, and considering the angle between them as 180 degree , i am bound to put cosθ = -1. So on doing ∞→R∫F.dr as ∞→R∫-Fdr, i get a positive answer which again should not be the case.
     
  6. Apr 25, 2016 #5
    I know nothing about calculus (yet)

    However how is the applied force and the displacement are in the same direction?
    The source charge is positive and the test charge is positive so the angle should be cos 180
    Which is -1
     
  7. Apr 25, 2016 #6
    OK, i don't really get your notation, use latex next time :D . I'll just write how you compute this and I hope it will be clear for you. So, if we define that the potential at infinity is zero, then : [itex]V(r)-V(\infty)=V(r)=- \int_{\infty}^{r} \vec{E} d\vec{l}[/itex] , by the definition. Here [itex]d\vec{l} = \hat{r} dr + r d\theta \hat{\theta} + rsin \theta d \phi \hat{\phi}[/itex]. Then:
    [itex]V(r)=- \int_{\infty}^{r} \vec{E} d\vec{l}=- \int_{\infty}^{r} E dr=- \int_{\infty}^{r} \frac{kq}{r^2} dr=-kq \int_{\infty}^{r} \frac{1}{r^2} dr=\frac{kq}{r}[/itex]

    Don't forget, always write [itex]d\vec{l}[/itex] in the most general possible way, like i wrote it in spherical coordinates, just do dot product and let your integral boundaries deal with minus and plus signs. Even if you are integrating from infinity to some r, your [itex]d\vec{l}[/itex] will be ''positive'', it will look from origin to infinity. Always. Your integral boundaries say whether you are going from infinity or to infinity. Hope it helped.
     
  8. Apr 25, 2016 #7
    Thank you again avalanche. I understand this derivation that u just gave me.But what intrigues me is the fact that a very primitive simple basic method of using the integral calculus to find out work done due to a variable force is not giving me the right result. I believe i am missing something or there is some mistake that i am making. I want to figure out why that method is not yielding me the result. Can you please help me find out where i am going wrong. Or what concept i applied wrong.
     
  9. Apr 25, 2016 #8
    Biker, The applied force, i believe should be in the same direction of the displacement. Here, the electrostatic force of repulsion should act away from the point say towards right considering the source charge at the origin and the point at a distance R to the right of it. Hence, an equal amount of applied force should be given in the opposite direction that is from right to left. And the test charge is moved from infinity to R again from right to left. Therefore the applied force and displacement are in the same direction.
     
  10. Apr 25, 2016 #9
    It is just when you get negative work in these condition (From the electric force) is like it takes the energy resulted of the applied force and store it as potential energy

    Sorry for interruption, Misunderstood the question.
     
  11. Apr 25, 2016 #10

    SammyS

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    II realize that you're rather new to PF and have not posted much. Also it's apparent that you are very anxious about this question.

    However, you need to show more patience here. This thread is very much similar to one you started some 5 hour prior to starting this thread. LINK Double posting a thread is against PF rules. I'm sure the Mentors will look at this.
     
  12. Apr 25, 2016 #11
    I understand, sincere apologies.
     
  13. Apr 26, 2016 #12

    robphy

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