Energy of parallel plate capacitor is defined as $$\frac{CV^2}{2}$$ where V is the potential difference between plates, but shouldnt we need to find the potential energy of each plate due to charge on it separetly and add them to get total energy
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$\begingroup$ Related: Energy of a system of conductors $\endgroup$– Alfred CentauriJun 27, 2019 at 12:13
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3 Answers
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. . . . we need to find the potential energy of each plate due to charge on it separately . . .
and having done that
and add them to get total energy
which neglects the work done in bringing the two plates closer together to form the capacitor.
What you are suggesting is very difficult to calculate.
You first need to evaluate the work done in assembling charge $+Q$ on a plate, then the work done in assembling charge $-Q$ on a plate which is very far away from the first plate and finally the work done in bringing these plates closer together to form the capacitor.
The net work done will be the potential energy stored by the capacitor.
The energy is stored in the electric field and if the electric field $E$ is constant then the energy stored per unit volume is $\frac 12 \epsilon E^2$ where $\epsilon$ is the permittivity of the medium.
So knowing what the electric field (and hence the energy stored) before the plates are brought together does not help as it is the electric field after the plates have been brought together which determines the energy stored by a capacitor.
For a parallel plate capacitor of area $A$ and plate separation $d$ with the electric field only existing between the plates of the capacitor the electric field is $\frac Vd$ where $V$ is the potential difference between the plates.
So the energy stored in the electric field between the plates of a capacitor is $\frac 12 \epsilon \left ( \frac Vd \right)^2 A d = \frac 12 \frac{\epsilon A}{d}V^2 = \frac 12 CV^2$
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$\begingroup$ But there is also an electric field outside the volume between the plates. Maybe that is negligible, but then I don't understand how this approximation can give the exact result? $\endgroup$– AzzinothJun 27, 2019 at 13:34
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1$\begingroup$ @Azzinoth, The calculation in Farcher's answer is for an ideal parallel plate capacitor, where the area is large enough to neglect any fringe electric field outside of the plates. If this fringe electric field is not negligible, it will slightly increase the capacitance, and thus the stored energy. Fringe electric field can be calculated for actual capacitors, though it often requires numerical methods for three dimensional geometries. $\endgroup$ Jun 27, 2019 at 14:16
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$\begingroup$ @amateurAstro Ah, so the approximation is in his formula for the capacity. I had overlooked that. Now it makes sense, thanks. $\endgroup$– AzzinothJun 27, 2019 at 15:27
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Remember potential energy is defined as the work done in assembling a charge configuration. That is if you have a set of charges arranged in space how much energy would be required to bring all of them together to their respective positions. Potential energy is therefore for a charge configuration, not for individual charges.
If you have two point charges at a distance $r$ the potential energy of the charge configuration is $$V_{energy} = \frac{q_1q_2}{4\pi\epsilon_0 r^2}$$
You don't take the individual potentials multiply with the corresponding charge and add them together.
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It is just the matter of definition,we define the electric potential energy of a parallel plate capacitor-system as the negative of work done by their mutual electrostatic forces when the system is released from rest.
answered Jun 27, 2019 at 5:58