1 Electrostatics 1 A Coulombs Law El Field

1. Electrostatics 1 A. Coulombs Law, El. Field, and Gauss’s Law Coulomb’s Law • Charges are measured in units called Coulombs • The force on a charge q at x from another charge q' at x': • The unit vector points from x' to x • We rewrite the unit vector as • For reasons that will make some sense later, we rewrite constant k 1 as • So we have Coulomb’s Law: • For complicated reasons having to do with unit definitions, the constant 0 is known exactly: • This constant is called the permittivity of free space

Multiple Charges, and the Electric Field • If there are several charges q'i, you can add the forces: • If you have a continuous distribution of charges (x), you can integrate: • In the modern view, such “action at a distance” seems unnatural • Instead, we claim that there is an electric field caused by the other charges – Electric field has units N/C or V/m • It is the electric field that then causes the forces

Gauss’s Law: Differential Version • Let’s find the divergence of the electric field: • From four slides ago: • We therefore have: • Gauss’s Law (differential version): • Notice that this equation is local

Gauss’s Law: Integral Version • Integrate this formula over an arbitrary volume • Use the divergence theorem: – q(V) is the charge inside the volume V • Integral of electric field over area is called electric flux Why is it true? • Consider a charge in a region • Electric field from a charge inside a region produces electric field lines q • All the field lines “escape” the region somewhere • Hence the total electric flux escaping must be proportional to amount of charge in the region

Sample Problem 1. 1 (1) A charge q is at the center of a cylinder of radius r and height 2 h. Find the electric flux out of all sides of the cylinder, and check that it satisfies Gauss’s Law • Let’s work in cylindrical coordinates • Electric field is: z • Do integral over top surface: r h h q h r • By symmetry, the integral over the bottom surface is the same

Sample Problem 1. 1 (2) A charge q is at the center of a cylinder of radius r and height 2 h. Find the electric flux out of all sides of the cylinder, and check that it satisfies Gauss’s Law • Do integral over lateral surface: z r h h q h r • Add in the top and bottom surfaces:

Using Gauss’s Law in Problems Gauss’s Law can be used to solve three types of problems: • Total electric flux out of an enclosed region – Simply calculate the total charge inside • Electric flux out of one side of a symmetrical region – Must first argue that the flux out of each side is the same • Electric field in a highly symmetrical problem – Must deduce direction and symmetry of electric field from other arguments – Must define a Gaussian Surface to perform the calculation – Generally use boxes, cylinders or spheres

Sample Problem 1. 2 A line with uniform charge per unit length passes through the long diagonal of a cube of side a. What is the electric flux out of one face of the cube? • The long diagonal of the cube has a length • The charge inside the cube is therefore • The total electric flux out of the cube is • If we rotate the cube 120 around the axis, the three faces at one end will interchange – So they must all have the same flux around them • If we rotate the line of charge, the three faces at one end will interchange with the three faces in back – So front and back must be the same • Therefore, all six faces have the same flux

Sample Problem 1. 3 A sphere of radius R with total charge Q has its charge spread uniformly over its volume. What is the electric field everywhere? • By symmetry, electric field points directly away from the center • By symmetry, electric field depends only on distance from origin Outside the sphere: • Draw a larger sphere of radius r • Charge inside this sphere is q(r) = Q • By Gauss’s Law, Inside the sphere: • Draw a smaller sphere of radius r • Charge inside this sphere is only • By Gauss’s Law, • Final answer:

Sample Problem 1. 4 An infinite line of charge has charge per unit length . What is the electric field everywhere? • Which direction does the electric field point? – Surely, it is radially away from the line • What can it depend on? – Only distance from the line • What Gaussian shape should we use? – Cylinder, length L, radius R, centered on the line • Gauss’s law says: • Enclosed charge is L • End caps have zero dot product • Lateral surface is a (rolled up) rectangle length L width 2 R • Solve for E:

1 B. Electric Potential Curl of the Electric Field • From homework problem 0. 1: • Generalize to origin at x': • Consider the curl of the electric field: • Using Stokes’ theorem, we can get an integral version of this equation:

Electric Field: Discontinuity at a Boundary Consider a surface (locally flat) with a surface charge • How does electric field change across the boundary? A L • Consider a small thin box of area A –L crossing the boundary • Since it is small, assume E is constant over top surface and bottom surface • Use Gauss’s Law on this small box • Charge inside the box is A • Since box is thin, ignore lateral surface • Consider a small loop of length L penetrating the surface • Use the identity • Ends are short, so only include the lateral part • So the change in E across the boundary is

The Electric Potential • In general, any function that has curl zero can be written as a gradient – Proven using Stokes’ Theorem • We therefore write: – is the potential (or electrostatic potential) – Unit is volts (V) It isn’t hard to find an expression for : • First note that • Generalize by shifting: • If we write: • Then it follows that:

Working with the Potential Why is potential useful? • It is a scalar quantity – easier to work with • It is useful when thinking about energy – To be dealt with later How can we compute it? • Direct integration of charge density when possible • We can integrate the electric field • It satisfies the Poisson equation: Solving this equation is one of the main goals of the next couple chapters • There is an ambiguity about , because it is an integral of the electric field – Constant of integration is ambiguous • Normally, resolved by demanding ( ) = 0

Sample Problem 1. 5 Find the potential and electric field at all points from a line charge with charge per unit length stretching from z = a to z = b along the z-axis. • Easiest to work in cylindrical coordinates: • Find the potential: • Maple: > Phi: = integrate(1/sqrt((zp-z)^2+rho^2), zp=a. . b); • To get electric field, use • Maple: > -expand(simplify(diff(Phi, z))); -diff(Phi, rho);

Sample Problem 1. 6 A sphere of radius R with total charge Q has its charge spread uniformly over its volume. What is the potential everywhere? • We already found the electric field • It makes sense that potential depends only on r • Relation between potential and electric field • So we have • Integrate this in the two regions: • We choose ( ) = 0, so • Want potential continuous at r = R, so • Put it together

Conductors • A conductor is any material that has charges in it that can move freely • If an electric field is present inside a conductor, then: – Charges will shift in response – These shifting charges will create electric fields – They will stop only when all electric fields are cancelled • Therefore, (perfect) conductors have E = 0 inside them • Recall that • Hence potential must be constant in a conductor • Consider Gauss’s law for any shape contained within the interior of a conductor • Since there is no electric field, there is no charge in the interior – There can be charge density on the surface of the conductor • Recall the discontinuity in the electric field at the surface • But it vanishes inside • Therefore, the electric field at the surface of a conductor is

Sample Problem 1. 7 a A neutral solid conducting sphere of radius R has a cubical cavity of side a inside it, with a charge q at its center. What is the electric field outside the sphere? • Consider any Gaussian surface surrounding the cavity • No electric field, so no charge inside • So the charge q must be balanced by charge –q from the conductor on the inside walls of the cavity • But there must be no net charge on the conductor, so charge + q on the outer surface of the conductor • This surface charge feels no force from the other charges, so it will distribute itself uniformly over the surface • This creates the same field as a point source at x = 0 – Nothing to do with where the charge is – Nothing to do with shape/size, etc. of cavity q –q +q

Sample Problem 1. 7 b A neutral solid conducting sphere of radius R has a cubical cavity of side a inside it, with a charge q outside. What is the electric field inside the cavity? q • The conductor is at constant potential, = 0. • The potential on the interior surface of the cavity is • There is no charge inside the cavity, so in the cavity, • As we will demonstrate shortly, the value on the boundary plus the value of the Laplacian is sufficient to determine the potential • I’m so good, I can get this solution by guess and check: – Check it yourself • Therefore, the electric field inside is • This has nothing to do with the shape of the cavity, the position of q, etc.

Potential and Potential Energy • Consider the formula: • Consider the relationship between energy and potential energy • By comparison, we see that • For example, for two charges q 1 and q 2 at x 1 and x 2: • If you have a lot of charges, – The i < j to avoid double counting • This is equivalent to • If you have a continuum of charges, this can be written as • Recall • We therefore have:

Sample Problem 1. 7 A sphere of radius R with total charge Q is being approached by a charge q and mass m with speed v from infinity. How fast must q be moving to make it to the center of the sphere? Q R • Use conservation of energy • We already know the potential • At infinity, the energy is • At the origin, the charge stops, so • Equating these, we have v q

1 C. Boundary Value Problems Definition of the Problem • Sometimes, we don’t know, don’t care, or don’t want to do the work of figuring out the potential everywhere • Sometimes, just want potential in a volume V – Which may be infinite • Poisson equation alone is not sufficient alone to determine potential • You need additional information about what happens to the potential at boundary • This information can take one of two types: – On Dirichlet boundary D, the potential is specified • Assume this happens somewhere perhaps x = – On Neumann boundary N, the normal derivative SD SN / n • Goal: find solution to these equations

Uniqueness of Solution? SD • • Could there be two solutions that satisfy all three equations? If 1 and 2 are both solutions of these, then define: It follows that Consider the integral: • • • Also, do the integral by divergence theorem On each surface, one of the factors vanishes The integrand is never negative Therefore, the only way to make it vanish is for the integrand to vanish Therefore U is constant If anywhere we have Dirichlet boundary conditions, then SN / n

Finding a “Simple” Solution SD • Consider the problem with a simple point source at x: • Let’s also let all the boundary conditions vanish: • We know the solution is unique: • If we had no boundaries, (except at ), solution would be • Let’s call the general solution (with zero boundaries): – These are called Green functions • This solution will have the properties: SN / n

How to Find or Check a Green’s Function SD • Green functions generally look like • The first term satisfies: SN / n • It follows that • The F term is just there to make the boundary conditions work out right • It is nonetheless usually hard to find If we think we have G, how do we check it? • Check that the Laplacian satisfies its equation • Check boundary conditions on Dirichlet boundaries (including G(x, ) = 0) • Check boundary conditions on any Neumann boundaries

Sample Problem 1. 8 Consider the half-space z > 0 with Dirichlet boundary conditions. Show that the function below is the correct Green’s function • Since x' = 0 is included in the space, we should have: • We also have the boundary z' = 0, at which we should have: • Finally, we have to check: • We are only interested in points within the allowed region – That means both z > 0 and z' > 0 • The second delta function can never vanish, so SD

Using the Green’s Function SD SN / n • G is the problem for a single point source • For a more general source, you would think you could then add the charges to get the general solution, so long as boundary conditions are still zero • Surprisingly, you can use G to get the general solution even when the boundary conditions aren’t zero • We will now demonstrate this

Potential from Green’s Functions (1) • Consider any two functions f and g of x • Consider the following identity: • Swap f and g: • Subtract them: • Integrate using divergence theorem: • Turn it around: – Normal derivatives: • Now, change variable x to x', let f be (x'), and let g be G(x, x'):

Potential from Green’s Functions (2) SD SN / n • Use the Laplacians above on left side: • Do the first integral • Break surface integral into Dirichlet part and Neumann part:

Sample Problem 1. 9 (1) Consider the half-space z > 0 conditions. Find the potential everywhere if there is no charge, and its boundary value is as given by • Since we are given potential, we need Dirichlet Green function: • The general solution (dropping the Neumann term): • The normal derivative is out of the region, so • We find: • No charge so SD

Sample Problem 1. 9 (2) Consider the half-space z > 0 conditions. Find the potential everywhere if there is no charge, and its boundary value is as given by SD • Can complete the integral by hand using a trig substitution: • Substituting in, we have

1 D. Capacitance and Energy Capacitance • Suppose you have conductors (and nothing else) in free space • Potential is constant on each • For each i, set Vi = 1 and Vj = 0 for j i, then there is some unique solution to this equation (x): • For arbitrary Vi the solution will be: • The charge on any conductor is then • Define the capacitance: – Units are Farads (F) • Then we see that • Can argue: S 1 S 2 V 1 V 2

Sample Problem 1. 10 What’s the capacitance of the Earth? • We have only one object, so we write • Treat the Earth as a spherical conductor of radius R = 6370 km • If we place a charge Q on it, the charge distributes itself uniformly over the surface • Electric field will be radially outward and depend only on r • By Gauss’s law, electric field will look like from a point charge: • The potential outside will also look like a point charge: • By continuity, the potential on the surface is then just • Solve for the capacitance: • Substitute numbers in:

Energy of a System of Conductors • Consider the formula for energy: V 1 • Only charge is on the surfaces, so V 2 • We therefore have:

Another Formula for Energy • • • Previous formula for potential energy: Recall: So we have: Now use the identity: We therefore have: • • On first term, use divergence theorem On second term, use E(x) = – (x) At infinity, E and both vanish We therefore have • Also equivalent to • Think of this expression as energy density: • Modern viewpoint: The energy density is in the electric field

1 E. Variational Method Minimization of Energy • Consider a problem with Dirichlet boundaries SD • We have an intuitive sense that systems try to minimize their energy • Can we find by minimizing this expression? • Not exactly, because we have to make sure is large when is large • For any function (x), consider the functional • We will pick so it has correct boundary conditions • We will demonstrate that if we pick = , we minimize this expression

Minimizes the Functional SD • Suppose that and differ by f(x): – Note that f vanishes on the boundary • Then we have • The last term is always positive • So this functional is minimized for the choice = • If we are close (f is small), we will get close (order f 2)

The Variational Method SD • Rather than picking a single function, pick a lot with the right boundary conditions • For each of them, find the functional: • Minimize it with respect to all the parameters • Then the approximation for the potential is

The Variational Method and Capacitance V • • • Suppose we have a capacitor with voltage V Set V = 1 and try to find the potential Use the variational approach Minimize The energy for a capacitor is • The energy is also given by • We therefore have • This is always an overestimate

Sample Problem 1. 11 Estimate the capacitance for a cube of side 2 a • Place the cube at the origin • Guess the functional form for potential – Must be 1 on the surface – Want it to fall off as 1/r at infinity • By symmetry, assume |x| is largest and x > 0 – Multiply by 6 to account for all the regions • We therefore have • Minimize with respect to the parameter b • Substitute back in

1 F. Relaxation Method Potential Can Be Found from Nearby Points • Consider the potential near a point x with no charge density and Dirichlet boundaries • Let’s work in 2 D (not sure why) • For a point offset in x-direction or y-direction will be approximately • Add these four points • We therefore have

Using a Grid • • Set up a rectangular grid: Label points by i and j Potential on boundaries is fixed In the interior, calculate using • Repeat until it converges • Comparably, you can use • Can show a more accurate method would be

Sample Problem 12 A square of side a has potential 0 on three sides and potential 1 on y = a. What is the potential at (x, y) = (a/4 , a/4) • Let’s set up a grid of size a/4 – Need 5 5 size grid • Set it up in an appropriate program – I used Excel • Type in all the boundary values • Put in one of the formulas • Recalculate repeatedly (F 9) until it converges • Increase grid size if you want