Electricity and Magnetism: Electricity

Electrical potential energy, U

The energy a charge has due to another charge being present.

U = \frac{1}{4 \pi ϵ_0}\frac{q_1 q_2}{r}

Electrical force, F

The force a charge experiences due to another charge being present.

\vec F = \frac{1}{4 \pi ϵ_0}\frac{q_1 q_2}{r^2}\hat r

Electric potential, V

The potential energy a charge at a point would have, per coulomb.

V = \frac{1}{4 \pi ϵ_0}\frac{q_1}{r}

Electric field, E

The electric force per positive unit charge at a point.

\vec E = \frac{1}{4 \pi ϵ_0}\frac{q_1}{r^2}\hat r

Permittivity

How easily electron clouds in a material absorb energy from an electric field, \epsilon=\epsilon_r\epsilon_0.

• \epsilon_r is relative permittivity and \epsilon_0 is permittivity of a vacuum, 1.

• \epsilon_r=1+\chi where \chi is the susceptibility to polarisation, where susceptibility of a vacuum is zero.

Properties of electric field lines

• They point in the direction a positive charge would travel.

• In systems with a net charge of 0, all field lines begin on a positive charge and end on a negative charge.

• The number of field lines per unit area through a surface perpendicular to the field lines is proportional to the magnitude of the field in that region.

How do you find the total electric field from a charged volume?

Work out the electric field from one tiny chunk of volume, then integrate over the total volume,

\vec E (\vec R) =\int_{V}\frac{1}{4 \pi ϵ_0}\frac{\vec R-\vec r}{|{\vec R - \vec r}|^3}\rho (\vec r) dV

How do you find the total electric field for discrete charges?

By applying the superposition principle: the total electric field due to a group of charges equals the vector sum of the electric fields of all the charges.

Electric flux, \Phi

Corresponds to the total number of field lines penetrating a surface

\Phi=\frac{\sum q_{enclosed}}{\epsilon_0}=\int_{S}\vec E \cdot \hat n\, da=\int_{S}\vec E \cdot \, d \vec a

Why Gauss’ law is valid for all surfaces and charge distributions?

The net flux through a surface doesn’t change with the surface shape or with where the charges are placed inside the surface.

Why is Gauss’ law independent of the distance from the charge when we draw the imaginary Gaussian surface?

The same net number of flux lines go through the surface no matter how far away we draw it.

Charge in a solid or hollow conductor that does not contain other charges

\vec E=0 inside a conductor at equilibrium → \Phi=0 through a Gaussian surface inside the conductor → using Gauss's law, will find that the net Q_{internal}=0

Electric field in a solid or hollow conductor in an external field

Electron clouds shift in response to external field → one side of the conductor becomes positive and the other negative, with all internal areas still neutral → this creates an opposing electric field in the conductor → this exactly cancels the external field, so E=0 inside the conductor.

Differential form of Gauss’ law

Used when E is not constant, and so cannot be taken outside of the integration.

\vec \nabla \cdot \vec E=\frac{\rho}{\epsilon_0}

If there are more flux lines going into A than out of B

The volume is acting as a sink of flux, so the divergence is negative; there is convergence.

If there are more flux lines coming out of B than in at A

The volume is acting as a source of flux, the divergence is positive.

If there are equal numbers of flux lines going into A as coming out of B

There is no net flux, and the divergence is zero.

Relation of force to energy

Force is the gradient of energy, F = - ∇ U

Electric dipole

A system with two charges of equal magnitude and opposite sign.

Electric dipole moment

A measure of how much the electrons have shifted over in a dipole, given by p=2qd, where 2d is the separation of the charges.

• This points in the direction a negative charge would go.

Polarisability

Polarisability (or susceptibility) describes how easy it is to push electrons to one side in a molecule and create a dipole.

Potential for a continuous charge distribution

V=\frac{1}{4\pi\epsilon_0}\int\frac{dq}{r}=\frac{1}{4\pi\epsilon_0}\int_{V}\frac{\rho(x_i,y_i,z_i)}{r_{ij}}d\tau

Potential for a discrete charges

Is equal to the work needed to bring the charges from an infinite separation to their final positions.

So for a collection of n charges q_1, q_2,...,q_n at positions \vec r_1, \vec r_2,...,\vec r_n:

U=\frac{1}{4\pi\epsilon_0}\sum_{all \,pairs}\frac{q_i q_j}{r_{ij}}=\frac{1}{4\pi\epsilon_0}\sum_{i,j=1,\,i\neq j}\frac{1}{2}\frac{q_i q_j}{r_{ij}}

Properties of a conductor in equilibrium

1. \vec E=0 everywhere inside a conductor, as the charges are stationary when in equilibrium.

2. If an isolated conductor carries a charge, all net charge resides on its surface (Gauss’ law)

3. The electric field just outside a charged conductor is perpendicular to the surface of the conductor. The tangential component of \vec E at the surface must be zero, otherwise the charges would move until this disappears.

Result of bringing two oppositely charged conducting plates together

The charges gradually move to the inside surfaces and as they do so the electric field on the outside goes to zero, while the field between the plates doubles.

Capacitor

Any combination of two conductors carrying charges of equal magnitude but opposite signs.

Capacitance, C

The charge that can be stored per volt of potential difference, C=\frac{q_1}{\Delta V}=\frac{\epsilon_0 A}{d}, units of farads.

Batteries vs capacitors

Batteries produce electrons at one terminal and remove at the other via chemical reactions. Capacitors don't make electrons, they only store them.

What kind of capacitor has the largest capacitance

One with the largest area and the smallest separation between plates.

(When not connected to a battery) Effect of increasing the separation in capacitor plates on potential difference and the electric field

The energy per charge increases so the potential difference increases, but the charge stored stays the same, so no change in the electric field.

(When connected to a battery) Effect of increasing the separation in capacitor plates on potential difference and the electric field

ΔV stays the same, as if the energy per charge stays the same then the work done to move a coulomb of charge from one plate to the other also stays the same. Therefore the force must decrease and so the electric field decreases.

Capacitors in series

There is the same charge on each plate → voltages add to give that of the battery → \frac{1}{C_{tot}}=\frac{1}{C_1}+\frac{1}{C_2}+…

Capacitors in parallel

The charge splits in proportion to the capacitances → there is the same voltage drop over each branch → equivalent to merging both negative plates, and both positive plates, so → C_{tot}=C_1+C_2+…

Energy stored in a capacitor

In a capacitor the average voltage drop is half the initial voltage drop, U_{cap}=\frac{1}{2}Q\Delta V_{initial}=\frac{1}{2}C(\Delta V_{initial})^2

Finding energy density of a capacitor

The total energy stored divided by the volume between the plates.

Electric polarisation of a dielectric, \vec P

The electric dipole moment per unit volume, \vec P=\frac{\Delta\vec p}{\Delta V}, which points from negative to positive.

Electronegativity

The more electronegative an atom, the more it will warp the electron cloud around it, creating an unequal distribution of charge.

Dielectric

An insulator, so an ideal dielectric has no free charges.

Nonpolar molecules in a capacitor

Molecules/atoms between the capacitor plates form a dielectric. The electrons move to one side of the molecule/atom so that it acquires a dipole moment and becomes polarised.

Polar molecules in a capacitor

Polar molecules between the capacitor plates already have a permanent dipole moment, but if they can rotate, they will align with an external field. They may also have an additional polarisation induced by this external field.

Charge capacitor, remove battery, then insert dielectric

Reduction in potential difference as the electric field between plates is partly cancelled out by molecules in the dielectric, so there is an increase in capacitance.

charge capacitor, keep battery in, then insert dielectric

Inserting the dielectric reduces the potential difference, which the battery then raises back (to the value on the battery) by increasing the charge held on each plate. As p.d. is the same but charge has increased, there is an increase capacitance.

What kind of dielectric increases capacitance the most?

The more susceptible the dielectric is to polarisation, the stronger this effect of increasing the capacitance (as the ‘cancelling’ effect of the dielectric is stronger).

Free charge vs bound charge

Bound charge is any that is part of a dipole, e.g. in a dielectric.

Free charge is any other charge, e.g. charge moving along a wire.

Resistors

Electrons undergo inelastic collisions and lose energy which is dissipated as heat in the resistor. Electrical potential drops across resistors and so resistors slow the current everywhere in the circuit.

Ohm’s law

For a constant temperature, current is proportional to potential difference, V=IR

Electron current vs conventional current

Conventional current flows from positive to negative, but electrons actually flow from negative to positive.

Current

The amount of charge that is flowing through a cross section of the circuit per second, I = \frac{dQ}{dt}

Conductors vs insulators

Conductors have an electron that will move on the application of very little electric force from a battery.

Insulators require a much larger potential difference from a battery to conduct.

Semiconductors usually act like insulators but can be made to act like conductors under certain conditions.

Current density, \vec J

The amount of charge that is flowing through a cross section of the circuit per second, per unit area.

\vec J is a vector in the direction of v_d, whose magnitude is the net amount of charge crossing a unit area perpendicular to the v_d in unit time:

\vec J=n\,q\,\vec v_d, where n is the number of charge carriers per unit volume,

Drift velocity

The electrons in a circuit move randomly with a slight bias in one direction due to the acceleration from the potential difference. Their net movement is slowly in that direction, at the 'drift velocity', v_d.

Relation between current and current density

I=\int_A\vec J \cdot d\vec a

Resistors in series

R_{tot}=R_1+R_2+…

Resistors in parallel

\frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}+…

Resistivity, \rho

Given by R=\frac{\rho L}{A}

Conductivity, \sigma

Defined as \sigma=\frac{1}{\rho}. This does not depend on the geometry of the conductor but is a property of the material from which the conductor is made.

Electrical power

The rate at which work is done, P=I\Delta V=I^2 R=\frac{(\Delta V)^2}{R}

Relation of current density to an electric field

Given by \vec J=\sigma \vec E

EMF

The work done per unit charge to move the charges against the electric field, \varepsilon = \frac{dW} {dq}

Kirchhoff’s loop rule

The sum of all potential differences around any closed circuit loop must be zero,

\sum_{\text{closed loop}}\Delta V=0

Which expresses the conservation of energy.

Kirchhoff’s junction rule

The sum of current entering any junction must be equal to the sum of current leaving it,

\sum I_{\text{in}}=\sum I_{\text{out}}

Which expresses the conservation of charge.

Where does energy go when a charge drops through a change in voltage in a resistor?

As the charge bumps into molecules in the resistor, it transfers energy and the resistor heats up.

Time-varying RC circuit equation

In a circuit with a capacitor and resistor in series (RC circuit), when charging the capacitor: q(t)=C\varepsilon\left(1-e^{\frac{-t}{RC}} \right) via Kirchhoff’s loop rule.

• RC is the time constant of the circuit.

• And current I(t) can be found by applying its definition, dQ/dt to this equation for charge.

Relation between electric field and potential

Given by \vec E=-\vec \nabla V

Poisson’s equation

\nabla² V=-\frac{\rho}{\epsilon_0}

Electrostatic equilibrium

When there are no electric fields or currents inside a conductor, and therefore no moving charges.

Gauss’ law

In a vacuum, the total flux leaving any closed surface is equal to the total charge contained within the surface divided by the permittivity of a vacuum,

\Phi= \int_{S}\vec E \cdot \, d \vec a=\frac{Q}{\epsilon_0}

Work done in moving charges

The potential energy of the charged object U is equivalent to the work done in bringing a unit charge from infinity to position \vec r against the electric field.

So integrate V=-\int_\infty^R \vec E \cdot \mathrm{d}\vec r.

Potential energy for a system of discrete charges

The work done to bring the charges from an infinite separation to their final positions.