At some point in lec. 7 of the lecture series, we have that the work done in moving the plate is given by 1/2 E^2 \epsilon_zero A x. We then have U = integral of the above dV. Is there a reason why we have missed out a minus sign here because I know F = -dU/dx?

Question

At some point in lec. 7 of the lecture series,  we have that the work done in moving the plate is given by 1/2 E^2 \epsilon_zero A x. We then have U = integral of the above dV.  Is there a reason why we have missed out a minus sign here because I know F = -dU/dx?

maheshmeghwal9 · Answer

because Prof. Walter Lewin hasn't considered Electric force there, (It is a conservative force).  
Actually he has taken external force there represented as Walter Lewin's Force which can't be considered as conservative therefor minus sign is avoided there.

& one more thing $$\color{blue}{\LARGE{F_{conservative}=- \frac{dU}{dx}}}$$

maheshmeghwal9 · Answer

& 
$$\color{blue}{\LARGE{F_{non-conservative}= \frac{dU}{dx}}}$$

caf123 · Answer

I have never seen such a non-conservative force described like that.  As far as I am aware of, you can only even start to define a potential if your force is conservative.  In Mathematics, the convention is to drop the minus and say $$\underline{F} = \underline{
abla} \phi, \phi$$ the scalar potential. Could you reference/point to a web link?

anonymous · Answer

The field isn't doing the work, the work is being done on the field.  You are applying a force against the electric field to increase the plate separation.

The energy stored in the field is increasing with increasing x so dU/dx = + 0.5 A E^2 / eps

The force applied to increase the separation is in the same direction as dU/dx so there's no minus sign missing in this case.