Static Electric Fields Static electric fields E ( x, y, z ) vary from - - PowerPoint PPT Presentation

Static Electric Fields

Static electric fields E(x, y, z) vary from one point (x, y, z) ∈ R3 to another but are constant with respect to time. Coulomb’s law of electrostatics Suppose a point charge of Q coul. is located at rs = (u, v, w) ∈ R3, and rt = (x, y, z) ∈ R3 is some other point (rt = rs). Then, the ele- ctric field at point rt due to the charge Q is the force exerted on a positive test charge of 1 coul. at rt, which is given by E(rt) = E(x, y, z) = Q 4πǫ0rt − rs2 (rt − rs) rt − rs

• unit vector

Here ǫ0 is the electrostatic permittivity of free space, which is given by ǫ0 = 8.854 × 10−12coul2/(Newton meter2).

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 1/22

Static Electric Fields (cont.)

Point charges are “singularities” and constitute rather unnatural

• bjects in the theory of electromagnetism.

It is much more usual to deal with charge that is diffusely spread through space and described by a charge density scalar field ρ. If (u, v, w) is a point within an infinitesimal cube dV ⊂ R3, then the total charge enclosed in the cube is given by dQ = ρ(u, v, w)dudvdw Thus, the total charge Q, diffusely spread throughout space ac- cording to the density ρ, is Q =

• R3 dQ =
• R3 ρ(u, v, w)dudvdw

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 2/22

Coloumb’s law for diffused charges

As a result of the charge dQ the electric field at (x,y,z) is the in- finitesimal vector given by

dE(x, y, z) = 1 4πǫ0 (x − u)i + (y − v)j + (z − w)k [(x − u)2 + (y − v)2 + (z − w)2]3/2 ρ(u, v, w)dudvdw,

with (u, v, w) being any point in dV . Our goal is to determine the total electric field at (x, y, z) as a result of all the charge contained within R3. Specifically,

E(x, y, z) =

• R3 dE(x, y, z) =
• 1

4πǫ0

• R3

(x − u)ρ(u, v, w) [(x − u)2 + (y − v)2 + (z − w)2]3/2 dudvdw

• i+

+

• 1

4πǫ0

• R3

(x − v)ρ(u, v, w) [(x − u)2 + (y − v)2 + (z − w)2]3/2 dudvdw

• j+

+

• 1

4πǫ0

• R3

(x − w)ρ(u, v, w) [(x − u)2 + (y − v)2 + (z − w)2]3/2 dudvdw

• k.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 3/22

Coloumb’s law for diffused charges (cont.)

Let us define Ψ(x, y, z) := − 1 4πǫ0

• R3

ρ(u, v, w) [(x − u)2 + (y − v)2 + (z − w)2]1/2 dudvdw Observe

∂ ∂x

• ρ(u, v, w)

[(x − u)2 + (y − v)2 + (z − w)2]1/2

• = −

(x − u)ρ(u, v, w) [(x − u)2 + (y − v)2 + (z − w)2]3/2 , ∂ ∂y

• ρ(u, v, w)

[(x − u)2 + (y − v)2 + (z − w)2]1/2

• = −

(y − v)ρ(u, v, w) [(x − u)2 + (y − v)2 + (z − w)2]3/2 , ∂ ∂z

• ρ(u, v, w)

[(x − u)2 + (y − v)2 + (z − w)2]1/2

• = −

(z − w)ρ(u, v, w) [(x − u)2 + (y − v)2 + (z − w)2]3/2 ,

for all (x, y, z).

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Coloumb’s law for diffused charges (cont.)

Recalling the gradient operator, we then have ∇Ψ(x, y, z) = ∂Ψ ∂x (x, y, z)i + ∂Ψ ∂y (x, y, z)j + ∂Ψ ∂z (x, y, z)k. Then, it is straightforward to see that E(x, y, z) = ∇Ψ(x, y, z), ∀(x, y, z) ∈ R3. Thus, an electric field arising from charge diffusely distributed through space according to a specified charge density field ρ is conservative. It immediately follows that the above electric field is irrotational, viz. ∇ × E = 0.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 5/22

Gauss’ law for static electric fields

Gauss’ law for static electric fields Suppose that E is the electric field arising from the charge density ρ. Then, for any region Ω ⊂ R3 with the closed outwardly oriented sur- face S as boundary, we have

• S

E · dA = 1 ǫ0

• Ω

ρ dV. According to Gauss’ law the total flux of E through S does not depend on the “outside” charge and is determined exclusively by the charge inside Ω. Gauss’ law is really just Coulomb’s law but stated in more esote- ric mathematical language.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 6/22

Differential form of Gauss’ law

Gauss’ law is a global statement since it gives an aggregate or net property of the electric field E. Let’s use the divergence theorem to rewrite Gauss’ law in a local

• r differential form which says something about E(x, y, z) at eve-

ry individual point (x, y, z). To this end, we first note that

• Ω

(∇ · E)dV =

• S

E · dA. From here, we obtain

• Ω

(∇ · E)dV = 1 ǫ0

• Ω

ρ dV.

• r equivalently
• Ω
• (∇ · E) − 1

ǫ0 ρ

• dV = 0.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 7/22

Differential form of Gauss’ law (cont.)

Since the above relation holds for any possible region Ω, it follows that (du Bois Reymond’s theorem) (∇ · E) = 1 ǫ0 ρ, which holds at each (x, y, z). Now recall that the electric field E related to a diffusively distri- buted charge is conservative, and hence there exists Ψ : R3 → R such that E = ∇Ψ. Therefore, ∇ · E = ∇ · (∇Ψ) = ∆Ψ, which suggests that ∂2Ψ ∂x2 + ∂2Ψ ∂y2 + ∂2Ψ ∂z2 = 1 ǫ0 ρ.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 8/22

Poisson equation

The equation ∆Ψ = (1/ǫ0)ρ is known as the Poisson equation, and it can be solved by very efficient means for virtually any ρ. Once Ψ is found, its related E can be found by means of diffe- rentiation (which is usually simpler than direct integration). Clearly, the solution of the above Poisson equation is given by

Ψ(x, y, z) = − 1 4πǫ0

• R3

ρ(u, v, w) [(x − u)2 + (y − v)2 + (z − w)2]1/2 dudvdw,

for all (x, y, z). Note that the above integral is a convolution integral.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 9/22

Static Magnetic Fields

The basic entity giving rise to a static magnetic field B is a static current density vector field J. We assume that J is defined on D := R3 and is static in the sense that at each (x, y, z), J(x, y, z) is constant with respect to time t. The basic experimental fact concerning magnetic fields B is the Biot-Savart law.

B(x, y, z) = µ0 4π

• R3

J(u, v, w) × [(x − u)i + (y − v)j + (z − w)k] [(x − u)2 + (y − v)2 + (z − w)2]3/2 du dv dw

for each (x, y, z) ∈ R3, where µ0 is the magnetic permeability of free space with the value µ0 = 4π × 10−7 henry/meter. Just as Coulumn’s law, Biot-Savart law is an “inverse square” law which is equivalent to Amp` ere’s circulate law and Gauss’ law for magnetic fields.

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Static Magnetic Fields (cont.)

Amp` ere’s circulate law and Gauss’ law for magnetic fields Suppose that B is the static magnetic field arising from the static cur- rent density J. Then, for each finite open surface S with boundary curve Γ, we have

• Γ

B · dr = µ0

• S

J · dA, and for each closed surface S we have

• S

B · dA = 0.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 11/22

Static Magnetic Fields (cont.)

In its local form, Gauss’ law for magnetic fields can be rewritten as ∇ · B = 0. As to Amp` ere’s circulate law, we have (by Stokes’ theorem)

• Γ

B · dr =

• S

(∇ × B) · dA, which suggests

• S

(∇ × B) · dA = µ0

• S

J · dA,

• r equivalently
• S

[(∇ × B) − µ0J] · dA = 0.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 12/22

Static Magnetic Fields (cont.)

Since

• S

[(∇ × B) − µ0J] · dA = 0. holds for each and every finite open surface S, one must have (by another du Bois Reymonds theorem) ∇ × B = µ0J. The above relation is the local form of Amp` ere’s circuital law. The curl of the magnetic field B at each (x, y, z) ∈ R3 is directly proportional to the vector J(x, y, z). In the case where J = 0 for some (x, y, z) ∈ R3, ∇ × B also can- not be identically zero. Therefore, the static magnetic field B arising from a static cur- rent density J cannot possibly be a conservative vector field.

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Remarks

Consider a general C1-vector field F : R3 → R3. If it’s true that F = ∇ × G for some C1-vector field G : R3 → R3 then ∇ · F = ∇ · (∇ × G) = 0, that is F = ∇ × G for some vector field G = ⇒ ∇ · F = 0. The converse is given by the following theorem. Theorem (Poincar´ e) Suppose that F : R3 → R3 is a C1-vector field. If (∇ · F)(x, y, z) = 0 for all (x, y, z) ∈ R3 then F is necessarily given by F(x, y, z) = (∇ × G)(x, y, z), ∀(x, y, z) ∈ R3, for some C1-vector field G : R3 → R3.

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Vector potential

Definition If a vector field F : R3 → R3 is given by F = ∇ × G, then the vector field G is called a vector potential of the vector field F. If F has a vector potential G, then F in fact has infinitely many vector potentials. To see this put ˜ G = G + ∇g for any C1-function g : R3 → R. Then, ∇ × ˜ G = ∇ × (G + ∇g) = ∇ × G + ∇ × ∇g = ∇ × G = F. In short, if G is a vector potential of F, then ˜ G = G + ∇g for any C1 function g : R3 → R is also a vector potential of F. That is why any solenoidal vector field F has infinitely many vector potentials.

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Remarks

Since the magnetic field B is solenoidal, we conclude that B = ∇ × A for some vector field A : R3 → R3. Any vector field A which satisfies the above relation is a magne- tic vector potential of the magnetic field B. if A is a magnetic vector potential of B, then for every C1-func- tion g : R3 → R, the vector field ˜ A := A + ∇g is also a magnetic vector potential of B.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 16/22

Computation of magnetic fields

The calculation of the magnetic field B in terms of J can be re- duced to the solution of three Poisson equations. To see that, we first recall Gauss’ law and Amp` ere’s law which are ∇ · B = 0 ∇ × B = µ0J. Now, let’s express B as B = ∇ × A, where A is a magnetic vector potential that satisfies ∇ · A = 0. Note that, since B has infinitely many vector potentials, it turns

• ut that we can always find a solenoidal one.

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Computation of magnetic fields (cont.)

Now we have ∇ × B = ∇ × (∇ × A) = ∇(∇ · A) − ∆A = −∆A. Then, Amp` ere’s law suggests that −∆A = µ0J. Since A = A1i + A2j + A3k, J = J1i + J2j + J3k, and ∆A = ∆A1i + ∆A2j + ∆A3k, we have three Poisson equations ∆Ai = −µ0Ji, i = 1, 2, 3, which can be solved either numerically or by direct integration.

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Coulomb gauge transformation

We still need to prove the existence of a solenoidal magnetic vec- tor potential. To this end, let ˜ A be any vector potential such that B = ∇ × ˜ A. Now, let f be any scalar vector field that satisfies ∆f = −∇ · ˜ A, and let A be defined as A = ˜ A + ∇f. A is also a magnetic vector potential of B, that is B = ∇ × A. Moreover, ∇ · A = ∇ · ( ˜ A + ∇f) = ∇ · ˜ A + ∇ · ∇f = ∇ · ˜ A + ∆f = 0. Thus, the magnetic vector potential A satisfies: ∇ · A = 0.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 19/22

Time Varying Fields

Law: Gauss’ law for time varying electric fields A time varying charge density ρ causes a time varying electric field E, and the fields ρ and E are related by (∇ · E)(x, y, z, t) = 1 ǫ0 ρ(x, y, z, t), for each t and each (x, y, z) ∈ R3. Law: Gauss’ law for time varying magnetic fields A time varying current density J causes a time varying magnetic field B which satisfies (∇ · B)(x, y, z, t) = 0, for each t and each (x, y, z) ∈ R3.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 20/22

Time Varying Fields (cont.)

Law: Faraday’s law of electromagnetic induction A time varying magnetic field B causes a time varying electric field E. Moreover, for each finite open surface S with boundary curve Γ, the fields B and E are related by

• Γ

E · dr = −

• S

∂B ∂t · dA. To rewrite Faraday’s law in a local form, we use Stokes’ theorem to obtain

• Γ

E · dr =

• S

(∇ × E) · dA, which leads to

• S

(∇ × E) · dA = −

• S

∂B ∂t · dA.

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 21/22

Time Varying Fields (cont.)

The later expression has the form of

• S
• ∇ × E + ∂B

∂t

• · dA = 0.

which holds for each and every finite open surface S as well as each instance t. By standard arguments, we therefore conclude that ∇ × E + ∂B ∂t = 0, for each t and each (x, y, z) ∈ R3. This relation is Faradays law of electromagnetic induction in local form. What about Amp` ere’s law?

Department of ECE, Fall 2014 ECE 206: Advanced Calculus 2 22/22