Capacitor

A capacitor that is an electrical device capable of storing and releasing electrical Charge in quick periods of times. It consists of two separate conducting slabs with an insulating material (or even a vacuum) between them. They are faster than batteries because the rate of energy transfer is not limited by the rate of chemical reactions in the battery. Useful applications include in a defibrillator.

Capacitance §

#todo Organise capacitance and capacitors #TODO talk in terms of resistance

Formula §

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$$C=\frac{{ϵ}_{0}A}{d}$$

• $A$ = Area of a single plate in the capacitor (in $m^2$)
• $d$ = Distance between plates (in $m$)

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Capacitance by Geometry

See how the formula uses only geometric measurements of area and distance, irrespective of charge? Capacitance can always be represented purely through geometric means such as area, length, volume, etc.

Electric Field §

In a capacitor with two plates of opposite charges, but equal in magnitude, $q$, the electric field between the plates is given by and separated by a vacuum:

Dielectric vs Vacuum

The following formulae apply when there is a vacuum separating the charged plates. In the real world, Dielectric materials are used to increase the Capacitance of the capacitor. See Dielectric Capacitor for this.

Formula §

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$$\vec{E}=\frac{\sigma }{{ϵ}_0}\hat{u}=\frac{q}{A{ϵ}_0}\hat{u}(\text{between plates})$$

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• $\vec{E}$ = Electric Field (in $N/C$ or $V/m$)
• $q$ = Magnitude of charge in each plate (in $C$)
• $\hat{u}$ = Unit vector pointing from the positive plate to the negative plate

Electric Potential §

Assuming we look at the potential from the positively charged plate to a point distance $x$ away from the + plate, i.e. parallel to $\vec{E}$. We define the positive plate to be $V=0$

Formula §

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$$\Delta V_{from+plate}=V_x-V_{+}=-\frac{\sigma }{{ϵ}_0}x=-\frac{q}{{ϵ}_0}x$$

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• $V$ = Electrical Potential (in $V$)
• $x$ = Distance from positively charged plate (in $m$)
  • In calculating the total potential of the capacitor, $x=d$ , where $d$ is the distance between plates.

Electrical Potential Energy §

Formula §

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$$\Delta U=\frac{1}{2}q\Delta V$$

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Derivations §

Electric Field §

Uses the same principles as that of Infinite Parallel Planes Of Charge. The main difference is that capacitors are constructed with the intention of having the plates have the opposite charges of the same magnitude:

Formula for electric field in parallel infinite charged plates:

In a capacitor, $q_2=-q_1$, and we’re interested in the fields between plates, so plugging it in:

$$\vec{E}=\frac{q_1-(-q_1)}{2{ϵ}_0}=\frac{2q}{2{ϵ}_0}=\frac{q}{e_0}$$

Electric Potential §

Electrical potential is defined as: We can use the top formula, since we already defined the electric field of our capacitor:

Let our initial point be at $x=0$ (at the + plate), to a point away from there.

$$\Delta V=-{\int }_i^f\vec{E}\cdot d\vec{L}=-{\int }_0^x\vec{E}dx$$

$$={\int }_x^0\frac{q}{{ϵ}_0}dx=-\frac{q}{{ϵ}_0}x$$

$$V_x-V_{+}=-\frac{q}{{ϵ}_0}x$$

If we define the +plate to be 0, then our potential decreases as we go away from the positive plate, and increases as we approach it.

However, if we define the -plate to be 0, which is usually done by convention, the potential increases as we go away from the positive plate.

Electrical Potential Energy §

Normally, we would expect the formula to hold the relationship between electric potential and electric potential energy, specifically, $\Delta U=q\Delta V$

Just a quick tangent, $\Delta U$ is electrical potential energy of a system, relative to a zero point. In a closed system of 2 point charges, $U_0$ is when the charges are infinitely far apart. However, we could also just as simply define $U_0$ to be when the the charges are almost touching, making the system neutral. Then any change in charge in the system is considered a change in $U$.

We’re going to use the second option here. Let $U=0$ when the plates when the plates are neutral. As the plates charge, we have a change from this position. To get the capacitor charged to a total charge of $Q$, we have infinite $\text{d}q$s moving from one plate to another.

The main problem here is that as the capacitor charges up, it also becomes harder to to move those $\text{d}q$s, because they start resisting against more charge. Let’s see why:

$$\Delta U=q\Delta V$$

Assume we started from the neutral plate position. Then $\Delta U=U_f-0=U$ and $\Delta V=V_f-0=0$

$$\text{d}U=\text{d}qV$$

$$U={\int }_0^{Q}V\text{d}q$$

We have $V=-\frac{q}{{ϵ}_{{}_0}}$ as we go from 0 to $d$, so $x=d$ ($d$ is distance between plates)

$$U={\int }_0^Q-\frac{q}{{ϵ}_0}d\text{d}q=-\frac{d}{{ϵ}_0}{\int }_0^{Q}q\text{d}q$$

$$=-\frac{d}{{ϵ}_0}[\frac{q^2}{2}{]}_0^Q=-\frac{d}{{ϵ}_0}(\frac{Q^2}{2})$$

Note that $V=-\frac{dQ}{{ϵ}_0}$ when $x=d$

$$U=\frac{Q}{2}V=\frac{1}{2}qV$$

$$\Delta U=\frac{1}{2}q\Delta V$$

#todo Show that this can be generalise for inductance in a conductor.

Resources §

• Potential Energy in a capacitor - Khan Academy
• Capacitors - OpenStax/Libretext