When a point charge is moved from one point say A to another point B in an electric field, then the work done by the field is stored as potential energy but what about the external work done?

Although the net work done is 0, what does that potential energy change mean to the overall system? Shouldn't it be 0?

I am a beginner. Please help me understand.


In short, the net work done on the charge is the change in kinetic energy. This is the Work-Energy principle.

In your example, the work done on the charge for an infinitesimal displacement is the component of the applied force along the direction of the displacement caused by said force:

$$ dW = \vec{F}\cdot d\vec{l} $$

In this case, to move the charge without imparting any leftover kinetic energy, you must exert an external force that exactly equal and opposite the electric force $q\vec{E}$, so $\vec{F} = -q\vec{E}$. For a finite displacement the total work is the sum over all the infinitesimal pieces of work, which just equals the difference in potential energy between the starting and ending points:

$$ W = \int_a^b \vec{F}\cdot d\vec{l} = -q\int_a^b \vec{E}\cdot d\vec{l} = q(V(b)-V(a))$$

So this change in potential energy is due to the force that you are applying. Hence, the external work done on the charge is the change in potential energy of the system if and only if the net work is zero.

  • $\begingroup$ Thank you for the answer! What about the work done by field? $\endgroup$ – Trilok Girish Kamagond May 31 at 4:37
  • $\begingroup$ I actually wanted to know which of the two causes change in potential energy? Another thing I want to know is, if work done by field is negative which is appearing as change in PE and work done by external force is -PE then the net PE is zero, but books tell that there is a PE change? $\endgroup$ – Trilok Girish Kamagond May 31 at 4:44
  • $\begingroup$ The work done by the field, $W_{\text{int}} = q\int_a^b \vec{E}\cdot d\vec{l}$, is minus the change in potential energy. $\endgroup$ – ultracoldgrl Jun 2 at 17:47
  • $\begingroup$ It is the external work which is moving the charge from $a$ to $b$, so the external work is what is actually causing this change in potential energy. If the net work is zero then the external work done on the charge is equal and opposite to the internal work done by the field, and there will be no change in kinetic energy. $\endgroup$ – ultracoldgrl Jun 2 at 17:55
  • $\begingroup$ Oh...now I get. Thank you so much for clearing $\endgroup$ – Trilok Girish Kamagond Jun 2 at 18:10

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