Lecture 4

Electric Potential

In this lecture you will learn:

• Electric Scalar Potential

• Laplace’s and Poisson’s Equation

• Potential of Some Simple Charge Distributions

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

Conservative or Irrotational Fields

Irrotational or Conservative Fields: r r Vector fields F for which ∇ × F = 0 are called “irrotational” or “conservative” fields r • This implies that the line integral of F around any closed loop is zero

r r ∫ F .ds = 0

Equations of Electrostatics: Recall the equations of electrostatics from a previous lecture: r ∇. εo E = ρ r ∇ × E = 0

⇒ In electrostatics or electroquasistatics, the E-field is conservative or irrotational

(But this is not true in electrodynamics)

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

1 Conservative or Irrotational Fields More on Irrotational or Conservative Fields: r • If the line integral of F around any closed loop is zero …..

r r ∫ F .ds = 0

r …. then the line integral of F between any two points is independent of any specific Path (i.e. the line integral is the same for all possible paths between the two points) r r ∫ F .ds = 0 rr rr ⎛ 2 r r ⎞ ⎛ 1 r r ⎞ ⇒ ⎜ ∫ F .ds ⎟ + ⎜ ∫ F .ds ⎟ = 0 path A r ⎜ r ⎟ ⎜ r ⎟ r2 ⎝ r1 ⎠path A ⎝ r2 ⎠path B rr rr ⎛ 2 r r ⎞ ⎛ 2 r r ⎞ rr ⇒ ⎜ ∫ F .ds ⎟ − ⎜ ∫ F .ds⎟ = 0 1 path B ⎜ r ⎟ ⎜ r ⎟ ⎝ r1 ⎠path A ⎝ r1 ⎠path B rr rr ⎛ 2 r r ⎞ ⎛ 2 r r ⎞ ⇒ ⎜ ∫ F .ds ⎟ = ⎜ ∫ F .ds⎟ ⎜ r ⎟ ⎜ r ⎟ ⎝ r1 ⎠path A ⎝ r1 ⎠path B

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

The Electric Scalar Potential - I The scalar potential:

Any conservative field can always be written (up to a constant) as the gradient of some scalar quantity. This holds because the curl of a gradient is always zero. r If F = ∇ϕ r Then ∇ × (F ) = ∇ × (∇ϕ ) = 0

r For the conservative E-field one writes: E = −∇φ (The –ve sign is just a convention)

Where φ is the scalar electric potential

The scalar potential is defined only up to a constant r If the scalar potential φ ( r ) gives a certain electric field then the scalar r potential φ ( r ) + c will also give the same electric field (where c is a constant)

The absolute value of potential in a problem is generally fixed by some physical reasoning that essentially fixes the value of the constant c

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

2 The Electric Scalar Potential - II

We know that: r E = −∇φ

This immediately suggests that: • The line integral of E-field between any two points is the difference of the potentials at those points r r2 rr rr 2 r r 2 r r r ∫ E .ds = ∫ ()− ∇φ .ds = φ()r1 − φ ()r2 r r r r r1 r1 1

• The line integral of E-field around a closed loop is zero r r r ∫∫E .ds = (− ∇φ ).ds = 0

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

The Electric Scalar Potential of a Point Charge Assumption: The scalar potential is assumed to have a value equal to zero at infinity far away from any charges r ∞ Point Charge Potential ds rr r q ˆ E = 2 r q 4πεo r

Do a line integral from infinity to the point r r where the potential needs to be determined 0 ∞ r r ∞ r r r r ∞ r r ∫ E .ds = ∫ ()− ∇φ .ds = φ()r − φ()∞ = φ()r ⇒ φ ()r = ∫ E .ds rr rr rr

r ∞ r r ∞ q φ()r = ∫ E .ds = ∫ dr r 2 r r 4π εo r q = 4π ε r o q φ()rr = 4π εo r

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

3 Electric Scalar Potential and Electric Potential Energy

The electric scalar potential is the potential energy of a unit positive charge in an electric field r • Electric force on a charge of q Coulombs = q E (Lorentz Law)

Potential energy of a charge q at = Work done by the field in moving the any point in an electric field charge q from that point to infinity

∞ r r ∞ r r r r Work done = ∫ F .ds = ∫ qE .ds = q[]φ()r − φ()∞ = qφ()r rr rr qφ(rr) Work done on unit charge = = φ()rr q

⇒ P.E. of unit charge = φ(rr) r ds ∞ rr

r ⇒ Potential energy of a charge of q Coulombs in electric field = qφ(r )

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

Poisson’s and Laplace’s Equation • It is not always easy to directly use Gauss’ Law and solve for the electric fields • Need an equation for the electric potential r Start from: ∇. εo E = ρ r Use: E = −∇φ

To get: ∇. εo ()− ∇φ = ρ

2 ρ ⇒ ∇ φ = − Poisson’s Equation εo

If the volume charge density is zero then Poisson’s equation becomes:

2 ∇ φ = 0 Laplace’s Equation

Poisson’s or Laplace’s equation can be solved to give the electric scalar potential for charge distributions

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

4 Potential of a Uniformly Charged Spherical Shell - I

2 Use the spherical coordinate system σ Coulombs/m a For a ≤ r ≤ ∞ :

∇2 φ = 0 Assume a solution:

1 ∂ ⎛ 2 ∂φ ⎞ r A ⇒ ⎜ r ⎟ = 0 φ()r = + F r 2 ∂r ⎝ ∂r ⎠ r F must be 0 so that the potential is 0 at r = ∞

For 0 ≤ r ≤ a :

∇2 φ = 0 Assume solution: 1 ∂ ⎛ ∂φ ⎞ ⇒ ⎜ r 2 ⎟ = 0 2 ∂r ∂r B r ⎝ ⎠ φ()rr = + D r

Potential must not become infinite at r = 0 so B must be 0

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

Potential of a Uniformly Charged Spherical Shell - II For 0 ≤ r ≤ a For a ≤ r ≤ ∞ r φ(r ) = D r A φ()r = a r σ ∂φ ∂φ A E ()r = − = 0 E ()r = − = r ∂r r ∂r r 2 Boundary conditions We need two additional boundary conditions to determine the two unknown coefficients A and D

(1) At r = a the potential is continuous (i.e. it is the same just inside and just outside the charged sphere) A D = a (2) At r = a the electric field is NOT continuous. The jump in the component of the field normal to the shell (i.e. the radial component) is related to the surface charge density

εo (Er out − Er in ) = σ ⎛ A ⎞ ⇒ εo ⎜ − 0⎟ = σ ⎝ a2 ⎠

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

5 Surface Charge Density Boundary Condition σ Suppose we know the surface normal electric field on just one side of a charge plane with a surface charge density σ E = ?? E1 2 Question: What is the surface normal field on the other side of the charge plane?

Solution:

• Draw a Gaussian surface in the form of a cylinder of area A σ piercing the charge plane

• Total flux coming out of the surface = εo (E2 − E1)A • Total charge enclosed by the surface = σ A E1 E2 = ?? • By Gauss’ Law: ε (E − E )A = σA o 2 1 A ⇒ εo ()E2 − E1 = σ

ε (E − E ) = σ This an extremely important result that relates surface normal o 2 1 electric fields on the two sides of a charge plane with surface charge density σ

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

Potential of a Uniformly Charged Spherical Shell - III For 0 ≤ r ≤ a For a ≤ r ≤ ∞

2 a r (4πσ a ) (4πσ a2 ) φ()r = φ()rr = σ 4π εo a 4π εo r

Sketch of the Potential:

φ()r

r a

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

6 Potential of a Uniformly Charged Sphere a la Poisson and Laplace In spherical co-ordinates potential can only be a function of r (not of θ or φ ) For a ≤ r ≤ ∞ : ρ Coulombs/m3 ∇2 φ = 0 Assume a solution: a 1 ∂ ⎛ 2 ∂φ ⎞ r A ⇒ ⎜ r ⎟ = 0 φ()r = + F 2 r r ∂r ⎝ ∂r ⎠ Work in spherical F must be 0 so that the For 0 ≤ r ≤ a : co-ordinates potential is 0 at r = ∞ ρ ∇2 φ = − εo Assume solution:

1 ∂ ⎛ 2 ∂φ ⎞ ρ ⇒ 2 ⎜ r ⎟ = − r B 2 r ∂r ⎝ ∂r ⎠ ε φ()r = + D + C r o r

homogenous particular parts solution ρ By substituting the solution in the Poisson equation find C C = − 6εo • Potential must not become infinite at r = 0 so B must be 0

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

Potential of a Uniformly Charged Sphere a la Poisson and Laplace For 0 ≤ r ≤ a For a ≤ r ≤ ∞ ρ ρ A φ()rr = D − r 2 φ()rr = a 6εo r

Boundary conditions We need two additional boundary conditions to determine the two unknown coefficients A and D

(1) At r = a the potential is continuous (i.e. it is the same just inside and just outside the charged sphere) ∂φ (2) At r = a the radial electric field is continuous (i.e. it is the E = − same just inside and just outside the charged sphere) r ∂r (1) gives: ρ 2 A ρ D − a = A = a3 6εo a 3ε ⇒ o

(2) gives: ρ 2 ρ A D = a 2εo a = 2 3εo a

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

7 Potential of a Uniformly Charged Sphere a la Poisson and Laplace For 0 ≤ r ≤ a For a ≤ r ≤ ∞ ρ

a ⎛ 4 3 ⎞ ρ ⎛ r 2 ⎞ ⎜ ρ π a ⎟ r ⎜ 2 ⎟ 3 φ()r = ⎜ a − ⎟ φ()rr = ⎝ ⎠ 2εo 3 ⎝ ⎠ 4π εo r

Sketch of the Potential:

φ()r

a r

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

The Principle of Superposition for the Electric Potential

Poisson equation is LINEAR and allows for the superposition principle to hold

• Suppose for some charge density ρ 1 one has found the potential φ 1

• Suppose for some other charge density ρ 2 one has found the potential φ 2

The superposition principle says that the sum ( φ 1 + φ 2 ) is the solution for the charge density (ρ1 + ρ2 ) A Simple Proof

2 ρ 1 2 ρ 2 (ρ + ρ ) ∇ φ1 = − + 2 = 1 2 ∇ φ2 = − ∇ ()φ1 + φ2 = − εo εo εo

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

8 Potential of a Charge Dipole Consider Two Equal and Opposite Charges z We are interested in the potential P at a distance r from the center of the pair in the plane of the d r+ r = r − cos()θ charges, where r >> d θ + 2 r r d Work in spherical co-ordinates + q − r = r + cos()θ − 2

d d Potential contributions from the cos()θ two charges can be added − q 2 algebraically q q φ()rr = − 4π εo r+ 4π εo r− q q = − ⎛ d ⎞ ⎛ d ⎞ 4π εo ⎜ r − cos()θ ⎟ 4π εo ⎜ r + cos()θ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ q d ≈ 2 cos()θ 4π εo r ECE 303 – Fall 2006 – Farhan Rana – Cornell University

Field of a Charge Dipole

r q d φ()r ≈ 2 cos()θ 4π εo r

r E = −∇φ()rr qd ˆ ˆ ≈ 3 ( 2cos()θ r + sin ()θ θ ) 4πεo r + q

− q

Same result for the E-field was obtained in the previous lecture by superposing the individual E-fields (rather than the potentials) of the two charges

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

9 Potential of a Line Charge

Consider an infinite line charge coming y out of the plane of slide λ • The electric field, by symmetry, has only a Coulombs/m radial component r x • Draw a Gaussian surface in the form of a cylinder of radius r and Length L perpendicular to the slide Work in cylindrical co-ordinates Using Gauss’ Law: εoEr (2π r L) = λ L λ ⇒ Er = 2π εo r ∂φ ∂ φ(rr) λ But Er = −∇φ = − ⇒ = − ∂r ∂r 2π εo r

λ ⎛ ro ⎞ Upon integrating from ro to r we get: φ()r − φ()ro = ln⎜ ⎟ 2π εo ⎝ r ⎠

Where ro is a constant of integration and is some point where the potential is known The problem is that this solution becomes infinite at r = ∞

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

Potential of a Line Dipole rr Consider two infinite equal and opposite line charges coming out of the plane of y slide r+ + λ r− − λ Coulombs/m Coulombs/m x d

Using superposition, the potential can be written as: λ ⎛ r ⎞ λ ⎛ r ⎞ φ()rr = ln⎜ o ⎟ − ln⎜ o ⎟ ⎜ ⎟ ⎜ ⎟ The final answer does not 2π εo ⎝ r+ ⎠ 2π εo ⎝ r− ⎠ depend on the parameter ro λ ⎛ r− ⎞ = ln⎜ ⎟ 2π εo ⎝ r+ ⎠

Question: where is the zero of potential?

Points for which r+ equals r- have zero potential. These points constitute the entire y-z plane

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

10 The 3D Superposition Integral for the Potential In the most general scenario, one has to solve the Poisson equation: ρ(rr) r r ∇2 φ()rr = − r − r ' εo rr We know that the solution for a point charge rr' sitting at the origin: q φ()rr = 4π εo r ρ

To find the potential at any point one can sum up the contributions from different portions of a charge distribution treating each as a point charge r r ρ(r ') dV ' = dx' dy' dz' φ()r = ∫∫∫ r r dV ' 4π εo r − r '

Check: For a point charge at the origin ρ(rr') = q δ 3 (rr') = q δ (x')δ (y')()δ z'

r 3 r r ρ()r ' q δ (r ') q q φ()r = ∫∫∫ r r dV ' = ∫∫∫ r r dV ' = r = 4π εo r − r ' 4π εo r − r ' 4π εo r 4π εo r

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

ECE 303 – Fall 2006 – Farhan Rana – Cornell University

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