Comparing energy lost by the battery & energy gained by the capacitor.

[nazmulhasanshipon]

Imagine the two terminal of a *parallel-plate capacitor* are connected to the two terminal of a battery with electric potential difference #V#. If the capacitance of the capacitor is #C#, and the area of each plate is $A$. In this process would the energy lost by the battery and the stored energy of the capacitor be the same or different? Please explain.

Someone pointed that the energy lost by the battery is #V=\frac{Qd}{\epsilon A}# (because the electric potential difference would be used to to raise the potential difference between the plates) and energy gained by the capacitor is #E=\frac{1}{2}QV#. And therefore they are different. But I doubt this since the energy should be conserved.

 

[nazmulhasanshipon]

Someone pointed that the energy lost by the battery is $V=\frac{Qd}{\epsilon A}$ (because the electric potential difference would be used to to raise the potential difference between the plates) and energy gained by the capacitor is $E=\frac{1}{2}QV$. And therefore they are different. But I doubt this since the energy should be conserved.

 

[DaveE]

If you are assuming that the battery is perfect voltage source (i.e. no losses, no inductance), then the question isn't well posed, it's not physically realizable. You would have to have an infinite current flow instantaneously. That is why the answers disagree between conservation of charge and conservation of energy.

For real circuits there are losses, which you can model with a resistor. In that case you will find that the energy lost is equal to the energy transferred.

If you built this out of superconductors, you would then need to include the inductance between the components, which is fundamental to moving charges, you can't get rid of it. In this case you will have a simple harmonic oscillator with infinite Q. The peak voltage on the capacitor will be twice the voltage step applied (a sinusoid) which accounts for the missing energy. When the capacitor voltage is equal to the battery voltage there will be current flowing through the inductor which will contain the missing energy in its magnetic field. That peak current will be the voltage step divided by the characteristic impedance $Z_o = \sqrt{\frac{L}{C}}$.

So, your circuit needs to also have inductance and maybe resistance too. With these assumptions there isn't a conflict between conservation of charge ($Q=CV$) and conservation of energy ($E=\frac{1}{2}LI^2+\frac{1}{2}CV^2+\int{RI^2}$).

 

[tech99]

Imagine the two terminal of a *parallel-plate capacitor* are connected to the two terminal of a battery with electric potential difference #V#. If the capacitance of the capacitor is #C#, and the area of each plate is $A$. In this process would the energy lost by the battery and the stored energy of the capacitor be the same or different? Please explain.

Someone pointed that the energy lost by the battery is #V=\frac{Qd}{\epsilon A}# (because the electric potential difference would be used to to raise the potential difference between the plates) and energy gained by the capacitor is #E=\frac{1}{2}QV#. And therefore they are different. But I doubt this since the energy should be conserved.

When the connection is made, energy is lost from the system due to the resistance of the wire.
The energy delivered by the battery is QV, but the energy gained by the capacitor, Ec, is less, being 1/2 CV^2.
C=Q/V so that Ec = 1/2 Q/V V^2 = 1/2 QV.
The loss resistance includes all forms of loss, including ohmic resistance, radiation and energy in a spark.

 

[Steve4Physics]

Someone pointed that the energy lost by the battery is $V=\frac{Qd}{\epsilon A}$

Q1. Can you explain why the above statement makes no sense?

Q2. If charge Q passes through a battery of emf V, how much energy has the battery converted (chemical to electrical) in terms of Q and V?

Q3. If a capacitor stores charge Q and has voltage V, how much energy does the capacitor store in terms of Q and V?

(You may assume ideal components: a battery with zero internal resistance, superconducting (zero resistance) wires and a lossless capacitor.)

 

[vanhees71]

You can make this quantitative. Take a circuit with a capacitor and resistor in series. Then in standard stationary approximation you have
$$Q/C+R i=U$$
or
$$Q/C+R \dot{Q}=U$$
The general solution is
$$Q(t)=c \exp[-t/(RC)]+C U.$$
The integration constant $c$ is determined by the intial condition $Q(0)=0$, leading to
$$Q(t)=C U \{1-\exp[-t/(RC)]\}.$$
For $t \rightarrow \infty$ you have $Q_{\infty}=C U$ as it must be, and the energy of the electric field in the capacitor is
$$E_{\text{el}}=Q_{infty} U/2=\frac{C}{2} U^2.$$
There's also heat energy due to the Ohmic loss in the resistor. The power is
$$P(t)=R i^2(t)=R \dot{Q}^2(t).$$
Now
$$\dot{Q}=\frac{C U}{R C} \exp[-t/(R C)]=\frac{U}{R} \exp[-t/(RC)]$$
and thus
$$E_{\text{heat}}=\int_0^{\infty} \mathrm{d} t P(t) =\frac{U^2}{R} \int_{0}^{\infty} \exp[-2 t/(R C)]=\frac{C}{2} U^2=QU/2.$$
So you store only 1/2 of the energy lost by the battery in the electric field within the capacitor. The other half is used to produce heat in the resistor.