# How to calculate the work of the electrostatic forces in a parallel-plate capacitor?

The expression of the energy stored in a parallel-plate capacitor is: $$U = \frac{e_0\cdot A \cdot V^2}{2d}$$ with $e_0$ the vacuum permittivity, $A$ the surface of the capacitor, $V$ the applied voltage and $d$ the distance between the two plates.

From what I understand, this energy equals the work of the electrostatic forces needed to get the plates from a zero separation (when they touch) to a separation d.

But now, let me try to actually calculate the work of the electrostatic forces. The electrostatic force applied to a plate is expressed as: $$F = \frac{e_0 \cdot A \cdot V^2}{2d^2}$$ So the work done by one plate on a distance d can be expressed as: $$W = \int_0^d{Fdx}=\int_0^d \frac{e_0\cdot A\cdot V^2}{2x^2}dx=\frac{e_0\cdot A \cdot V^2}{2}\int_0^d\frac{dx}{x^2}$$ which obviously diverges as the integral of $\dfrac{1}{d^2}$ is $-\dfrac{1}{d}$ and we need to calculate it for $d = 0$...

So what is wrong with my reasoning? How to calculate the work of the electrostatic force so as to obtain the actual expression of the electrostatic energy $U$?

• Your initial assumption is wrong. The work is the work required to transfer the charge onto the plates against the EMF produced by the charge already on the plates. The result should be $\tfrac{1}{2}CV^2$. The plate separation remains constant. – John Rennie Aug 19 '15 at 6:15

I think your approach isn't wrong; however in your calculations you're making the assumption that the potential difference between plates, $V$, is constant: What remains constant is the charge on each plate. So the equation becomes: $$W=\int_0^d {{q^2} \over {2\epsilon_0A}} \;\mathrm{dx}={{q^2d} \over {2\epsilon_0A}}$$ Since $C=\epsilon_0A/d$ we obtain $$W={q^2\over{2C}}={1\over2}CV^2$$ When you implicitly assume $V$ is constant, electric field when $d=0$ becomes infinite, hence the integral does not converge.

As John Rennie marked, the result should be $\tfrac{1}{2}CV^2$.

Let me deduce this for you;

Let's start with an uncharged condenser & by some means you remove one an electron from one plate & transfer it to the other plate. You have to do hardly any work to transfer the first electron but as you gradually continue the process, the field that is emanated due to the transfer of negative charge on the other plate & the increase of positive charge on the first plate hinders you to transfer any further negative charge. Thus, as the process goes on, you have to do greater work than the previous time.

Now, the work you do against the field is stored as potential energy in the system as $U$.

Deduction of $U = \frac{CV^2}{2}$:

Suppose at a given instant, a charge $q^*$ has been transferred from one plate to the other one. The potential difference that will develop after the transfer is $V^* ={q^*\over C}$. . If you then pick again some infinitesimal charge $dq$ & transfer it, then you have to an extra work of $dw = V^* dq = {q^*\over C} dq$ . So, the total work done against the electrostatic force to bring a negative of charge of magnitude $q$ to the second plate is : $$W = \int_0^W dw = \frac{1}{C} \int_{0}^{q} q^* dq = \frac{q^2}{2C} = \frac{(CV)^2}{2C} = \frac{CV^2}{2}\implies U = \frac{\epsilon_0\cdot A \cdot V^2}{2d}$$.

Second approach to deduce $U = \frac{1}{2}CV^2$:

I saw OP has used the thought-experiment of displacing one plate from other; so I think this derivation needs to get mentioned so as to shun all confusions.

Let there be a condenser having surface area of each plate as $A$. Suppose the first plate is deprived of $-Q$; so it is now positively charged of magnitude $Q$. Electric field due to this positive plate is $E_{+} = \frac{q^*}{2A\epsilon_0}$(considering the plate sufficiently large to neglect the end points). Therefore, force on the negatively-charged plate due to the positively-charged one is $$F = -Q E_{+} \implies |F| = \frac{Q^2}{2A \epsilon_0}$$. the plates attract one another with a force equal to this.

Now, consider the two plates to be infinitesimally far away i.e. $d \approx 0$. Suppose one plate is moved slowly to a distance $s$ keeping the other fixed. The force at any instant by the fixed plate on the other is $F = \frac{Q^2}{2A\epsilon_0}$. In order to move the plate with constant velocity(no increment of kinetic energy) against the attractive electrostatic force, you have to apply the same force against the electrostatic force in the opposite direction.

The work done by you which will be stored as $U$ in the system is $$W = F\int_0^s ds = \frac{Q^2 d}{2A\epsilon_0} = \frac{Q^2}{2C} = \frac{\epsilon_0\cdot A \cdot V^2}{2s}$$.

The calculation is not right, but the result is true. You need infinite work to pull apart two oppositely charged infinite plates, they have each infinite charge!

The electric field due to one infinite plate is $\frac{\sigma}{2 \epsilon_0}$. The force on an area $A$ of the second plate will thus be $\frac{- \sigma^2 A}{2 \epsilon_0}$, which is infinite for an infinite area.

To solve all our troubles we should then assume a finite area! But we are still going to work with the $E$ field of the infinite plate.

The work done on a plate of finite area $A$ is then $\frac{\sigma^2 A}{2 \epsilon_0} d$ in order to bring it from $0$ to $d$.

This is exactly the energy stored $C V^2 \over 2$ once you set $C = \frac{\epsilon_0 A}{d}$ and $V = \frac{Q}{C} = \frac{\sigma A}{C}$.

From what I understand, this energy equals the work of the electrostatic forces needed to get the plates from a zero separation (when they touch) to a separation d.

That's not the ordinary understanding and, from the electrical circuit perspective, the energy stored equals the work done by the external circuit separating electric charge in the capacitor (moving electric charge from one plate to the other via the external circuit).

To supplement the answers already given, I will show yet another approach from the electric circuit perspective. Start with the governing equation for an ideal capacitor

$$Q = C \cdot v_C$$

where it is understood that the plates of the capacitor have equal and opposite charge $Q$, and the voltage across (potential difference) is $v_C$.

The time rate of change of Q equals the electric current $i_C$ into the more positive terminal:

$$\dot Q = i_C = C \cdot \frac{dv_C}{dt}$$

In a circuit context, the power delivered to the capacitor is just the product of the voltage across and current 'through'

$$p = v_C \cdot i_C = \frac{Q}{C}\cdot C\frac{\dot Q}{C} = \frac{Q^2}{C}\frac{dQ}{dt}$$

But the power delivered to the capacitor is just the time rate of change of work done on the capacitor thus

$$\frac{dW_C}{dt} = \frac{Q^2}{C}\frac{dQ}{dt}$$

or

$$dW_C = \frac{Q^2}{C}dQ$$

thus

$$W_C = \frac{1}{2}\frac{Q^2}{C} = \frac{1}{2}Cv^2_c$$

The work done by the external circuit is stored as electric potential energy in the capacitor and so this is the energy stored by the capacitor.

This result is general. In the specific case that the capacitor is a parallel plate capacitor, we have that

$$C = \frac{\epsilon_0 A}{d}$$

thus, for a parallel plate capacitor, the energy stored is

$$W_{C_{||}}= \frac{1}{2}\frac{\epsilon_0 A}{d}v^2_C$$