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

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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

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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