# Numb3rs: Capacitance

Capacitance is the ability of a Capacitor to store charge.

## Capacitor A Capacitor is an electrical component that can store energy in an electric field. They consist at least of two electrical conductors separated by a dielectric (insulator), although many variations can be found.

A potential voltage difference between the two fields will create a static electric field develops across the dielectric. Positive charge will collect on one of the plates and negative charge on the other one.

## Charge

$q = C * V$

the charge stored equals to the Capacitance of the object and the Voltage across the object.

## Energy

To derive an expression for how much energy is stored in a capacitor we consider the infinitesimal small voltage due to the infinitesimal little charge $$y$$ $C * dV = dq$

Energy is defined as: $E = q * V$

The Energy for every small amount of voltage can therefore be expressed as: $dE = q * dV$

At this point we can take our Capacitance expression and plug it in. $dE = q \frac{dq}{C}$

Then we can add all these infinitesimal small Energies up to get the total amount of energy: $\int dE = \int q \frac{dq}{C}$ $E = \frac{1}{2} \frac{q^2}{C} = \frac{1}{2} \frac{(C V)^2}{C} = \frac{1}{2} C V^2$

## Capacitance

The Capacitance between two plates can be derived as followed

### Electric Field

The Electric Field between two plates: $E = \frac{\sigma}{\epsilon_0} = \frac{q}{A} \frac{1}{\epsilon}$

### Voltage

The Voltage between two plates: $V = E * d = \frac{d * q}{A\epsilon_0}$

### Capacitance $q = CV$ $C = \frac{q}{V} = \frac{A \epsilon_0}{d*q} q = \frac{A \epsilon_0}{d}$

The Capacitance between two plates: $C = \frac{\epsilon_0 A}{d}$

## Examples

In these examples we are trying to find:

• $$E$$, the Electric Field
• $$\Delta V = \int_A^B E(r) dr$$, the difference in Voltage
• $$C = \frac{q}{V}$$, the Capacitance

### Cylinder

__IMAGE_PLACEHOLDER__

#### Electric Field

$\Phi = \oint \vec{E} d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$ $\oint \left| \vec{E} \right| * \left| d\vec{A} \right| * \cos (\theta i) = \frac{Q_{enc}}{\epsilon_0}$

The $$\cos$$ term will fall away since the Field is perpendicular to the area. $\oint \left| \vec{E} \right| * \left| d\vec{A} \right| = \frac{Q_{enc}}{\epsilon_0}$

The Area will just add up to the total Area since again all the small Area’s are at the same angle to the Electric Field. $A \oint \left| \vec{E} \right| = \frac{Q_{enc}}{\epsilon_0}$ $2 \pi r l * \oint \left| \vec{E} \right| = \frac{Q_{enc}}{\epsilon_0}$

Therefore the Electric Field will be: $\oint \left| \vec{E} \right| = \frac{Q_{enc}}{\epsilon_0 2 \pi r l}$

#### Voltage Difference

Voltage is defined as: $V = E * d$

Therefore we can define the change in Voltage as $\Delta V = \int_A^B E dr$

We can plug in our result from earlier to get $\Delta V = \int_A^B \frac{Q_{enc}}{\epsilon_0 2 \pi r l} dr$

We can factor out all the constants to get $\Delta V = \frac{Q_{enc}}{\epsilon_0 2 \pi l} \int_A^B \frac{1}{r} dr$ $\Delta V = \frac{Q_{enc}}{\epsilon_0 2 \pi l} \left( \ln{B} - \ln{A} \right)$ $\Delta V = \frac{Q_{enc} \ln{\frac{B}{A}} }{\epsilon_0 2 \pi l}$

#### Capacitance

Capacitance is defined as $C = \frac{Q}{V}$

Plug and play $C = \frac{Q_{enc}}{ \frac{Q_{enc} \ln{\frac{B}{A}} }{\epsilon_0 2 \pi l} }$

Solving Magic $C = \frac{\epsilon_0 2 \pi l}{ \ln{\frac{B}{A}} }$

### Sphere

__IMAGE_PLACEHOLDER__

#### Electric Field

$\Phi = \oint \vec{E} d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$ $E(r) * 4 \pi r^2 = \frac{Q_{enc}}{\epsilon_0}$

Therefore the Electric Field will be: $E(r) = \frac{Q_{enc}}{\epsilon_0 * 4 \pi r^2}$

#### Voltage Difference

Same steps as for the cylinder can be applied: $\Delta V = \frac{Q_{enc}}{\epsilon_0 * 4 \pi} * \int_A^B \frac{1}{r^2}$

The integral can simply be solved by usage of the power rule on $$r^{-2}$$. $\Delta V = \frac{Q_{enc}}{\epsilon_0 * 4 \pi} * \left( \frac{1}{A} - \frac{1}{B} \right)$

#### Capacitance

Capacitance is defined as $C = \frac{Q}{V}$

Plug and play $C = \frac{Q_{enc}} {\frac{ Q_{enc} \left( \frac{1}{A} - \frac{1}{B} \right) } {\epsilon_0 4 \pi} } = \frac{\epsilon_0 4 \pi}{ \left( \frac{1}{A} - \frac{1}{B} \right) } = \frac{\epsilon_0 4 \pi}{ \left( \frac{B-A}{AB} \right) } = \frac{\epsilon_0 4 \pi \left( AB \right) }{ B-A }$

• My Notes

## The Numb3rs Series

Numb3rs used to be its own subsection on this site, where I wrote math & science tutorials. Unfortunately this never came to frution so I integrated ths articles with the rest of the blog. If you want to check out the other Numb3rs posts, you can get an overview here.

## By Cecil Wöbker I do science during the day and develop or design at night. If you like my work, hire me.

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