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

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

Sources

• My Notes