Suppose we have a grounded conducting sphere (potential of sphere = 0) and we bring a charged particle to its surface.

Since work done by us would be equal to the change in potential energy of the charge, it will be equal to zero, As potential at infinity = potential on sphere = 0 therefore change in potential energy = 0.

But this is not possible as when we move the charge, it will induce a negative charge on the sphere, thus for charge to move without acceleration, a negative force (with respect to our displacement) would be required. Thus we would do negative work all over the path and work done can never be zero.

Am I wrong in assuming that the work done = change in potential in this case? Because only that would make sense. Any help would be appreciated.

  • 1
    $\begingroup$ See this question:physics.stackexchange.com/questions/139448/…. $\endgroup$ – Jan Bos Sep 11 '16 at 11:20
  • $\begingroup$ So in this case, work done is NOT equal to change in potential energy? If yes then can someone elaborate on when the work done equals change in potential energy $\endgroup$ – Rishabh Sep 11 '16 at 11:25
  • 1
    $\begingroup$ No it still is but there is a problem with how you set the reference at infinity and the grounded sphere alike . You have to pick one. $\endgroup$ – Jan Bos Sep 11 '16 at 11:53
  • $\begingroup$ Ok let's suppose I take the potential at infinity to be zero, then what value should I give to the sphere? $\endgroup$ – Rishabh Sep 11 '16 at 12:02
  • 1
    $\begingroup$ Per the answer in the below (using an image charge in the sphere) you can solve for potential outside and on the surface of the sphere. I suppose it will depend on the radius of the sphere and the separation. $\endgroup$ – Jan Bos Sep 11 '16 at 12:15

Lets proceed like this. The surface of the sphere S1 being grounded is at a potential we are labelling as V1 and an imaginary system boundary S2 at infinity is as V2(imagine a colossal spherical surface S as the boundary, this S2 is basically the sphere of influence of the grounded sphere.) WHen we start bringing a positive charge, it is initially outside S2(no interaction initially). So initially, the boundaries of my system S1 and S2 are at potential V1 and V2(fixed). After i bring the charge onto the conductor. The potential V1 should've changed, but 'grounding' essentially means a technique to keep the potential constant no matter how large a finite charge is added. Technically, the test charge should also be small enough for this assumption. So now V1 is unchanged. Obviously V2 is unchanged too, and therr is no charge in between S1 and S2(my system). So the Neumann boundary conditions for thr laplace equation are valid. The field E between S1 and S2 must therefore be unchanged. Since there is no change in $E^2$dv, the net field energy is same. Hence no work has been done on the system in consideration.


The potential at a point is the work done in order to bring a unit positive charge from infinity to that point, without any acceleration. Electric potential is a location-dependent quantity that expresses the amount of potential energy per unit of charge at a specified location.

For a conservative field, we have by definition

$$\vec{E}=-\nabla V$$

which means the electric field at a point is equal to the negative gradient of the absolute potential at that point. This assumes the fact that the work done by the external agency in moving a charged particle through a uniform static electric field is the same for all paths, otherwise dependent only on the initial and final positions.

So, the electric field is due to some charge, say $q$ and the potential is observed on some other charge $Q$, a unit test charge, which by definition, should not alter or distort the electric field of the charge $q$.

This means, simply you cannot take into account the negative charge formed on the sphere. To solve such problems, you need to use the method of images


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.