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Lecture 16 - Midterm 1 Review

Coulomb's law

Magnitude of the force between two charges:

$F=k\frac{Q_{1}Q_{2}}{r^{2}}$

where $k=8.99\times 10^{9}\mathrm{Nm^{2}/C^{2}}$

The proportionality constant $k$ can be written $k=\frac{1}{4\pi \epsilon_{0}}$. $\epsilon_{0}$ is the permittivity of free space, $\epsilon_{0}=8.85\times10^{-12}\mathrm{C^{2}/Nm^{2}}$.

Note that in any given case a charge exerts an equal and opposite force on another charge to the force that the other charge exerts on it, and thus satisfies Newton's third law.

Like all forces, electric forces are vectors.

Electric Fields

In order to consider the general effect that a charge or distribution of charges will have on charges in their vicinity we can introduce the concept of an electric field.

The electric field $\vec{E}$ is defined as the force $\vec{F}$ a small positive test charge with charge $q$ would experience, divided by that charge, i.e.

$\vec{E}=\frac{\vec{F}}{q}$

The SI units of electric field are $\mathrm{N/C}$ or $\mathrm{V/m}$.

The magnitude of the electric field due to a single point charge $Q$ at a distance $r$ from the charge is

$E=k\frac{Qq}{r^{2}}\frac{1}{q}=k\frac{Q}{r^{2}}=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}$

Given the field $\vec{E}$ at a point in space we can work out the force $\vec{F}$ on an arbitrary charge $q$ using

$\vec{F}=q\vec{E}$

A positive charge experiences a force in the same direction as the field, a negative charge feels a force in the opposite direction to the field.

Electric field from charge

The field from a point charge is a vector $\vec{E}$ which has magnitude

$E=\frac{1}{4\pi\epsilon_{0}}\frac{Q}{r^{2}}$

and when $Q$ is positive points radially outwards from the charge.

The field from an element of charge, which also points radially outwards from the charge element has magnitude

$dE=\frac{1}{4\pi\epsilon_{0}}\frac{dq}{r^{2}}$

If we want to find the field directly from a distribution of charge we can integrate over the charged distribution, but we have to be careful to do take in to account the vector nature of the field!

We only tackle problems where there is cancellation of components of the field from different parts of the charge distribution.

Field distributions due to continuous charge distributions

We define an infinitesimal charge element $dQ$ which is a distance r from the point at which we want to know the field. The field due to this charge element $dE$ is

$dE=\frac{1}{4\pi\epsilon_{0}}\frac{dQ}{r^{2}}$

The electric field is

$\vec{E}=\int d\vec{E}$

Integration over a charge distrbution

Suppose we want to integrate over a distribution of charge. What does $\int dq$ really mean?

For a line of charge $\int dq = \int \lambda\,dx$
For a sheet of charge $\int dq = \int \sigma\,dA$
For volume of charge $\int dq =\int \rho\,dV$

Depending on the symmetry of the problem we might choose to integrate over Cartesian coordinates or polar coordinates, so for example,

for a rectangular sheet of charge $\int dq = \int \sigma\,dA=\int \int \sigma \,dx\,dy$
for a circular sheet of charge $\int dq = \int \sigma\,dA=\int \int \sigma r\,dr\,d\theta$

Volume integrals in terms of shells

Often we come across integrals of spherical or cylindrical volumes (as these have easier symmetry than other volumes…)

In cases where $\rho$ is dependent either constant or only depends on r

For spherical volume the integral $\int\int\int\rho\,dV=\int\rho A dr=\int\rho 4\pi r^{2}dr$

For cylindrical volumes the integral $\int\int\int\rho\,dV=\int\rho A dr=\int\rho 2\pi r l dr$

Volume integrals in terms of slices

When we have a volume with planar symmetry we instead can divide our volume in to slices of area A, so for a generic prism with area $A$ in which the charge density is uniform in the directions along $A$,

$\int\int\int\rho\,dV=\int\rho A dl$

Electric Flux

If we consider a uniform electric field $\vec{E}$ passing through a flat area $A$ we can define the electric flux as $\Phi_{E}=EA\cos\theta$

Area vector

We would like to define the electric flux in terms of a vector dot product. In order to do this we represent the area by a vector with magnitude equal to the area and direction normal to the surface.

Now we can go from

$\Phi_{E}=EA\cos\theta$

to

$\Phi_{E}=\vec{E}.\vec{A}$

Gauss's Law

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$

Gauss's law will often give us a better way to evaluate the field for a distribution of charge, because on the right hand side we only need to evaluate the total amount of charge.

The most important step is to choose an appropriate Gaussian surface which reflects the symmetry of the charge.

We should remember that the Gaussian surface must be a closed surface, and so it may be composed of more than one surface.

If possible we want to choose the surfaces so that the field on each surface is uniform and we don't have to actually any integration over the surface.

It's important to remember that flux has a sign. It is positive when it is leaving the volume, and negative when entering (as the $\vec{A}$ always points out).

Spherical symmetry

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$

If the field points out radially from the center of the sphere

$\oint\vec{E}.d\vec{A}=\oint E\,dA$

and if it is uniform over the surface

$\oint E\,dA=E \oint dA=E4\pi r^{2}$

As for the enclosed charge

$Q=\int\int\int\rho\,dV=\int\rho A dr=\int\rho 4\pi r^{2}dr$

Cylindrical symmetry

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$

If we consider the line of charge to be infinitely long then the field points straight out from the line, so the flux through the end circular surfaces is zero and for the curved surface

$\oint\vec{E}.d\vec{A}=\oint E\,dA$

as the field is uniform over the surface

$\oint E\,dA=E \oint dA=E2\pi rl$

To find the enclosed charge

$Q=\int\int\int\rho\,dV=\int\rho A dr=\int\rho 2\pi r l dr$

Planar symmetry

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$

If the plane is infinite then the flux through the curved surfaces is zero and we only need to be concerned about the flux through the flat prism surfaces of area $A$. The flux through those two faces will not always be the same, so in general if we define a field through the left face $E_{L}$ and the right face $E_{R}$

$\Phi_{E}=\oint\vec{E}.d\vec{A}=E_{L}A+E_{R}A$

$Q=\int\int\int\rho\,dV=\int\rho A dl$

or when there is sheet of charge $\sigma$

$Q=\int \sigma\,dA=\sigma A$

Charged metal sphere

Consider a hollow metal sphere with a net positive charge.

Because of the spherical symmetry for any spherical Gaussian surface of radius r

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\oint EdA=E\oint dA=EA=E4\pi r^{2}$

and Gauss's law gives us

$\Phi_{E}=\oint\vec{E}.d\vec{A}=\frac{Q}{\varepsilon_{0}}$

Any spherical Gaussian surface we draw inside the sphere of charge does not include any charge and so the net electric flux through the sphere must be zero.

By contrast if we draw a spherical gaussian surface outside the charge we can show that the field outside the sphere is $E=\frac{Q}{4\pi\varepsilon_{0}r^{2}}$

If there is charge inside the sphere, then there will be charge both on the inner and outer surface of the sphere.

Electric potential and field

The electric potential and field are easily related.

$U_{b}-U_{a} = -\int_{a}^{b}\vec{F}\cdot\,d\vec{l} $

$V_{b}-V_{a}=\frac{U_{b}-U_{a}}{q}=-\int_{a}^{b}\vec{E}\cdot\,d\vec{l} $

When the field is not uniform we can obtain the potential from the field by integration. Note again that the path of the line integral has not been specified: we are free to choose the most convenient path to facilitate the calculation.

Potential due to a Charge Distribution

We also need to be able to calculate the potential due to a charge distribution

$V_{b}-V_{a}=-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}$

$V=\frac{1}{4\pi\varepsilon_{0}}\int\frac{dq}{r}$

In general, use the first formula when Gauss's law easily gives the field, and the second formula where the distance $r$ is easily written in terms of the spatial variables that define $dq$ and the resulting integral is not too hard to do.

We can get the field from the potential from

$\vec{E}=-\hat{i}\frac{\partial V}{\partial x}-\hat{j}\frac{\partial V}{\partial y}-\hat{k}\frac{\partial V}{\partial z}$

Potential due to a point charge

If we want to find the change in potential in going from a distance $r_{a}$ to a distance $r_{b}$ from a single point charge we need to evaluate the integral

$V_{b}-V_{a}=-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}$

We can recall that the field is

$E=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r^{2}}$

If we integrate along a path which is directly radially outward $\vec{E}$ and $d\vec{l}$ are in the same direction and

$-\int_{r_{a}}^{r_{b}}\vec{E}.d\vec{l}=-\frac{Q}{4\pi\varepsilon_{0}}\int_{r_{a}}^{r_{b}}\frac{1}{r^{2}}dr=\frac{Q}{4\pi\varepsilon_{0}}[\frac{1}{r}]_{r_{a}}^{r_{b}}=\frac{Q}{4\pi\varepsilon_{0}}(\frac{1}{r_{b}}-\frac{1}{r_{a}})$

It is common to define the potential as being zero at $\infty$, and if we do this here (setting $r_{a}=\infty$) the potential as a function of the distance $r$ from a point charge is

$V=\frac{1}{4\pi\varepsilon_{0}}\frac{Q}{r}$

Potential directly from charge

It can be easier to find the potential due to multiple charges or a distribution of charges than it is to find the field.

We denote the negative charge as $A$ and the positive charge as $B$ and we can then say that the potential is given by

$V=\frac{1}{4\pi\varepsilon_{0}}(-\frac{Q}{R_{A}}+\frac{Q}{R_{B}})$

Potential due to a continuous distribution of charge

From the previous example we can conclude that for $n$ point charges each which produces a potential $V_{i}$ at point A the potential at point $A$ will be

$V_{A}=\sum\limits_{i=1}^{n}V_{i}=\frac{1}{4\pi\varepsilon_{0}}\sum\limits_{i=1}^{n}\frac{Q_{i}}{r_{ia}}$

Now if we want to instead consider a continuous charge distribution

$V=\frac{1}{4\pi\varepsilon_{0}}\int\frac{dq}{r}$

Relationship between Field and Potential

$\Delta U = -\int_c\vec{F}\cdot\vec{dl}\Longrightarrow \vec{F} = -\vec{\nabla}U$.

When we want to find the gradient vector field we use an operator $\vec{\nabla}$ which is read as del

$\vec{\nabla}=\hat{i}\frac{\partial}{\partial x}+\hat{j}\frac{\partial}{\partial y}+\hat{k}\frac{\partial}{\partial z}$

When applied to a scalar field we can use the word grad, so we would say “$E$ is equal to minus grad $V$”

$\vec{E}=-\nabla V $

$\vec{E}=-\hat{i}\frac{\partial V}{\partial x}-\hat{j}\frac{\partial V}{\partial y}-\hat{k}\frac{\partial V}{\partial z}$.

If one moves along an equipotential line, $dV$ is zero. This means that the $\vec{\nabla}V$ is perpendicular to that line, which shows why electric field lines are perpendicular to equipotentials.

Divergence of the electric field

The divergence of a vector field gives a measurement of whether it acts a source or sink of flux. We can see this by applying the divergence theorem to Gauss's law.

$\oint_{A} \vec{E}\cdot d\vec{A}=\int_{V}\nabla\cdot \vec{E}\, dV \to \int_{V}\nabla\cdot \vec{E}\, dV=\frac{Q}{\varepsilon_{0}} $

The total amount of charge may also be written as a volume integral so that

$\int_{V}\nabla\cdot \vec{E}\, dV=\frac{1}{\varepsilon_{0}}\int_{V}\rho\, dV$

where $\rho$ is the charge density. As the integration volume on each side of the equation is equivalent and arbitrary we can say that

$\nabla\cdot \vec{E}=\frac{\rho}{\varepsilon_{0}}$

This is referred to as the differential form of Gauss's law.

The Parallel Plate Capacitor Configuration

This consists of two metal plates with area $A$ carrying equal and opposite charges $Q$. The field due to one of these sheets of charges can be found from Gauss's law

$E=\frac{Q}{\varepsilon_{0}A}$

The potential due to this sheet of charge is

$V=-\frac{Qd}{\varepsilon_{0}A}$

Here we have defined the zero of potential as the position of the positive sheet of charge.

The Definition of Capacitance

For capacitors it makes most sense to define the voltage as the potential difference across the capacitor and give it a positive sign, ie.

$V=\frac{Qd}{\varepsilon_{0}A}$

Charges will not spontaneously separate across a capacitor, we need to apply a potential to the capacitor to move charges from one plate to another. The amount of charge $Q$ separated when a given voltage $V$ is applied is obtained from the equation above

$Q=\frac{\epsilon_{0}A}{d}V$

The capacitance $C$ is the amount of charge separated per volt applied

$C=\frac{Q}{V}$

and we can see for a parallel plate capacitor

$C=\frac{\epsilon_{0}A}{d}$

The unit for capacitance is the farad, $\mathrm{F}$ (named after Michael Faraday) which is equivalent to $\mathrm{\frac{C}{V}}$

Dielectric in a field

Since the dipoles reduce the electric field strength inside a dielectric, one can write generally:

$E_{dielectric} = \frac{1}{K}E_{vacuum}$

where $K$ is a constant (specific to each material) greater than 1.

At the boundary of the dielectric the boundary condition is that

$E_{vacuum}=\frac{P}{\varepsilon_{0}}+E_{dielectric}$.

For most dielectric materials we can say that the polarization is given by $\chi \varepsilon_{0} E_{dielectric}$ where $\chi$ is the dielectric susceptibility.

The quantity $K=1+\chi$ is called the dielectric constant of the material.

As there is no way for dipoles to occur in vacuum it has susceptibility $\chi=0$, giving $K=1$

Air has very little susceptibility and so $K$ for air is just a little bit greater than 1

For a dielectric material where $K>1$ the field is less in the dielectric than in the air. A key characteristic of a dielectric is the partial exclusion of the external field from the interior of the material.

Dielectric in a capacitor

$E_{dielectric}=\frac{Q}{\varepsilon_{0} A}-\frac{P}{\varepsilon_{0}}=\frac{Q}{\varepsilon_{0} A}-\chi E_{dielectric}$

$KE_{dielectric}=\frac{Q}{\varepsilon_{0} A}$

$E_{dielectric}=\frac{V}{d}$

$K\frac{V}{d}=\frac{Q}{\varepsilon_{0} A}$

$C=\frac{Q}{V}=K\frac{\varepsilon_{0}A}{d}$

In general, the capacitance of a configuration with a dielectric is related to the same geometry with a vacuum gap:

$C_{dielectric} = KC_{vacuum}$.

Capacitors in circuits

Circuit diagrams are a useful way to represent an electrical circuit. We introduce two symbols for capacitors and batteries so that we can draw circuits with capacitors in series and in parallel.

Capacitors in parallel

When capacitors are connected in parallel the potential across each capacitor is the same. Each acquires a charge determined by it's capacitance and the total charge is the sum of these charges

$Q=Q_{1}+Q_{2}+Q_{3}=C_{1}V+C_{2}V+C_{3}V=(C_{1}+C_{2}+C_{3})V$

$Q = C_{eq}V$

We can then consider an equivalent capacitance which is the sum of the capacitors

$C_{eq}=C_{1}+C_{2}+C_{3}$

Capacitors in series

When capacitors are connected in series it is the charge on each capacitor which is the same, as the region between the capacitors, which is considered to be an ideal conductor must remain overall electrically neutral and cannot have a field.

$Q=C_{eq}V$

The total voltage must equal the voltage across the battery

$V=V_{1}+V_{2}+V_{3}$

For each capacitor $Q=C_{i}V_{i}$

$\frac{Q}{C_{eq}}=\frac{Q}{C_{1}}+\frac{Q}{C_{2}}+\frac{Q}{C_{3}}$

$\frac{1}{C_{eq}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}+\frac{1}{C_{3}}$

Current

If an amount of charge $\Delta Q$ flows in a time interval $\Delta t$ the average current in this time is defined as

$\bar{I}=\frac{\Delta Q}{\Delta t}$

The instantaneous current is the limit as $\Delta t \to 0$

$I=\frac{dQ}{dt}$

The unit for current is the Ampere $\mathrm{A}$ which is equivalent to $\mathrm{\frac{C}{s}}$, and is named after Andre Ampere. This is usually abbreviated as an amp.

Even though in metallic wires it is negative electrons that flow the convention for current is for a flow of positive charges, so current flows from high to low potentials, even though when electrons carry the current they are actually moving from low to high potential.

Ohm's law

Ohm's law state that

$V=IR$

It should be noted that Ohm's law is not valid for all materials, and typically the resistance $R$ depends on factors such as temperature. For example, as a filament light bulb heats up it's resistance will increase.

Electric Power

When an infinitesimal amount of charge $dq$ moves through a potential $V$ the change in the potential energy of the charge is

$dU=Vdq$

The work done on the charge is $dW=-dU=-Vdq$, the work done by the charge is equal to $dU$.

We can recall that power is the rate at which work is done, and so

$P=\frac{dU}{dt}=\frac{dq}{dt}V=IV$

The above is a general relationship, for an ohmic conductor only, where $V=IR$

$P=VI=I^{2}R=\frac{V^{2}}{R}$

We should be careful here, the power dissipated in a given component needs to be calculated using only the potential change from one end to other!

Circuit elements

We have now discussed the properties of batteries, capacitors and resistors.

We will now look at how the behave when connected together in circuits.

For ease of communication, there are a standard set of symbols used for circuit diagrams.

Resistors in Series

If we recall our formula for resistance

$R=\rho\frac{l}{A}$

we can expect that if we connect two resistors in series the total resistance should be the sum of the resistances.

The current $I$ which flows through the circuit is the same everywhere.

The potential in each resistor is given by Ohm's law $V=IR$.

The total potential is the sum of the potentials across the resistors.

$V=V_{1}+V_{2}+V_{3}=IR_{1}+IR_{2}+IR_{3}$

We can define an equivalent resistance for the 3 resistors

$R_{eq}=R_{1}+R_{2}+R_{3}$

Resistors in Parallel

When resistors are connected in parallel the current splits itself up between the various branches

$I=I_{1}+I_{2}+I_{3}$

As the potential across each resistor is the same $V$

$I_{1}=\frac{V}{R_{1}}$     $I_{2}=\frac{V}{R_{2}}$     $I_{3}=\frac{V}{R_{3}}$

The total current is

$I=\frac{V}{R_{eq}}$

where $R_{eq}$ is the equivalent resistance of the 3 resistors.

As $\frac{V}{R_{eq}}=\frac{V}{R_{1}}+\frac{V}{R_{2}}+\frac{V}{R_{3}}$

$\frac{1}{R_{eq}}=\frac{1}{R_{1}}+\frac{1}{R_{2}}+\frac{1}{R_{3}}$

Kirchoff's Rules

Kirchoff's junction rule comes from conservation of charge and says that for any junction in a circuit the sum of all currents entering the junction must equal the sum of all currents leaving the junction.

Kirchoff's loop rule comes from conservation of energy and states that the changes in potential around any closed loop of a circuit must be zero.

The way to use these rules is to use them set up a sufficient number of equations based on the junctions and loops in a circuit to solve for the required unknowns.

Solving the circuit

$I_{1}+I_{2}=I_{3}$
$\mathcal{E}_{1}-I_{3}R_{3}-I_{1}R_{1}=0$
$\mathcal{E}_{2}-I_{3}R_{3}-I_{2}R_{2}=0$

Suppose we know the values of the resistors and the emfs and wish to find the currents.

$\mathcal{E}_{1}-I_{3}R_{3}-I_{1}R_{1}=0\to$
$\mathcal{E}_{1}-I_{3}R_{3}-(I_{3}-I_{2})R_{1}=0\to$
$\mathcal{E}_{1}-I_{3}(R_{3}+R_{1})+I_{2}R_{1}=0 $

$\mathcal{E}_{2}-I_{3}R_{3}-I_{2}R_{2}=0\to I_{2}R_{1}=(\mathcal{E}_{2}-I_{3}R_{3})\frac{R_{1}}{R_{2}}$

$\mathcal{E}_{1}-I_{3}(R_{3}+R_{1})+(\mathcal{E}_{2}-I_{3}R_{3})\frac{R_{1}}{R_{2}}=0$

$\mathcal{E}_{1}R_{2}-I_{3}(R_{3}R_{2}+R_{1}R_{2}+R_{3}R_{1})+\mathcal{E}_{2}R_{1}=0$

$I_{3}=\frac{\mathcal{E}_{1}R_{2}+\mathcal{E}_{2}R_{1}}{R_{3}R_{2}+R_{1}R_{2}+R_{3}R_{1}}$

$\mathcal{E}_{1}-I_{3}R_{3}-I_{1}R_{1}=0\to\mathcal{E}_{1}-\frac{\mathcal{E}_{1}R_{2}+\mathcal{E}_{2}R_{1}}{R_{3}R_{2}+R_{1}R_{2}+R_{3}R_{1}}R_{3}-I_{1}R_{1}=0$

$I_{1}=\frac{1}{R_{1}}(\frac{\mathcal{E}_{1}R_{3}R_{2}+\mathcal{E}_{1}R_{1}R_{2}+\mathcal{E}_{1}R_{3}R_{1}-\mathcal{E}_{1}R_{2}R_{3}-\mathcal{E}_{2}R_{1}R_{3}}{{R_{3}R_{2}+R_{1}R_{2}+R_{3}R_{1}}})=(\frac{\mathcal{E}_{1}R_{2}+\mathcal{E}_{1}R_{3}-\mathcal{E}_{2}R_{3}}{{R_{3}R_{2}+R_{1}R_{2}+R_{3}R_{1}}})$

$I_{2}=(\frac{\mathcal{E}_{2}R_{1}+\mathcal{E}_{2}R_{3}-\mathcal{E}_{1}R_{3}}{{R_{3}R_{2}+R_{1}R_{2}+R_{3}R_{1}}})$

RC circuit: Mathematical Solution

From Kirchoff's loop rule

$\mathcal{E}-I(t)R-\frac{Q(t)}{C}=0$

Writing this in terms of charge

$\mathcal{E}-\frac{dQ}{dt}R-\frac{1}{C}Q=0$

and then

$\mathcal{E}C-Q=\frac{dQ}{dt}RC$

$\frac{dQ}{\mathcal{E}C-Q}=\frac{dt}{RC}$

At time $t=0$, $Q=0$ and at time $t=t$, $Q=Q$

$\int_{0}^{Q}\frac{dQ}{\mathcal{E}C-Q}=\int_{0}^{t}\frac{dt}{RC}$

$[-\ln(\mathcal{E}C-Q)]_{0}^{Q}=[\frac{t}{RC}]_{0}^{t}$    ⇒    $\ln(\frac{\mathcal{E}C}{\mathcal{E}C-Q})=\frac{t}{RC}$    ⇒    $\ln(\frac{\mathcal{E}C-Q}{\mathcal{E}C})=-\frac{t}{RC}$    ⇒    $\frac{\mathcal{E}C-Q}{\mathcal{E}C}=e^{-\frac{t}{RC}}$

$Q=\mathcal{E}C(1-e^{-\frac{t}{RC}})$

The voltage on the capacitor is given by $V_{C}=\frac{Q}{C}$

$V_{C}=\mathcal{E}(1-e^{-\frac{t}{RC}})$

$I=\frac{dQ}{dt}=\frac{\mathcal{E}}{R}e^{-\frac{t}{RC}}$

phy142kk/lectures/14.txt · Last modified: 2015/03/02 07:08 by kkumar
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