In my course, it is written that :

If we have a charge $q$ in an electric field $\vec{E}$,then, the work done by $\vec{E}$ if it moves from $a$ to $b$ in space is given by : $$W_{a,b}=\int_{a}^{b}\vec{E}\cdot \vec{dl}$$ the work I must do to move the charged particle from $a$ to $b$ is given by : $$W_{a,b,me}=-W_{a,b}$$

Now, we consider a capacitor (figure) that consists of two conductors separated by a distance $d$. Suppose that the surface charge density is uniform and that $\sigma_1=\sigma_2=\sigma$. By using the fact that a physics student move one of the conductor by a distance $dx$, I would like to compute the force $\vec{F}$ that is applied on this conductor. So, I know that the student is doing a work by moving the capacitor up with a distance $dx$ : $$W_{st}=F_{st}dx$$ where $F_{st}$ is the magnitude of the force exerted by the student. Then, if I take a small charge $dq$ on the conductor $1$, two forces are applied on $dq$, considering that gravity does not have any impact here. Then, according to my course, I have : $$W_{st}=-W_{elec}$$ where $W_{elec}$ is the work done by the electrical force due to the field $\vec{E}$ between the two conductors... BUT, in the solution of this problem, it is written that : $$dE_{total}=W_{elec}+W_{st}$$ and I don't understand this ...

Is it because, in the course, the only force that is applied on the particule is the one due to the electric field ? Any help would be appreciated, Diagramm of the capacitor

  • $\begingroup$ (a) The first equation in your course should read $$W_{a, b}=q \int_a^b \vec E.\vec {d \mathscr{l}}$$ (b) Like you, I don't understand your last equation. $W_{elec}$ is presumably the $W_{a, b}$ already defined, and is the loss in electric PE of $q$ in the $\vec E$ field. This loss in PE is equal to the work done ON the student, that is $-W_{st}$. So $W_{elec} + W_{st}=0$ So your last equation gives $dE_{total}=0$. This is no doubt telling us that the system consisting of $q$, $\vec{E}$ and student doesn't gain or lose energy! Not very enlightening. There are, in my opinion, easier methods... $\endgroup$ – Philip Wood Dec 7 '19 at 17:40
  • $\begingroup$ In fact, in the solution, it is said that $dE_{total}$ is non zero ! it is equal to $d(E_{cin}+E_{pot}+E_e)$ (kinetic energy of the system + potential energy due to gravity + electrostatic energy of the capacitor), and it is said that $E_{pot}$ and $E_{cin}$ are constant, but not $E_e$, so that $dE_e=W_{elec}+W_{st}$ ... $\endgroup$ – Dicordi Dec 7 '19 at 17:52
  • $\begingroup$ and thanks for the $(a)$ ! $\endgroup$ – Dicordi Dec 7 '19 at 17:52
  • $\begingroup$ You probably know this, but the electric force on one plate of a parallel plate capacitor carrying charges $±Q$ is $\tfrac12 QE$. The $\tfrac 12$ arises because only $\tfrac12 E$ is due to the $other$ plate! Alternatively equate the work done as the plates come together by a distance $\Delta x$ to the capacitor's loss in stored energy. $\endgroup$ – Philip Wood Dec 7 '19 at 18:54

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