Lecture 8 Electrostatic Potential & Gauss Law

8/2/2019 Lecture 8 Electrostatic Potential & Gauss Law

1/11

EECS 117

Lecture 8: Electrostatic Potential & Gauss Law

Prof. Niknejad

University of California, Berkeley

University of California, Berkeley EECS 117 Lecture 8 p. 1/


2/11

Electric Dipole

x

y

z

r

r+

r+q

q

An electric dipole is two equal and opposite charges qseparted by a distance d

Consider the potential due to an electric dipole at pointsfar removed from the dipole r d

(r) =q

4r+ + q

4r =q

4 1

r+ 1

r



3/11

Law of Cosines

Vector summation is fundamentally related to a triangle.Note that we can think of adding two vectors r and zd/2

as forming r+ = r + zd/2

The length of a vector is given by |a|2 = a aWe can therefore express r+ in terms of r and d

r+ =r+ r+ =

(r +

d

2z) (r + d

2z)

Or simplifying a bit and recalling that r d

r+ = r1 + d2r2

+d

r

cos

r1 + d

r

cos



4/11

Back to Potential Calculation

So we have

1

r+

=1

r

11 + dr cos

1

r1

d

2r

cos and

1r+

1r

1 + d2r

cos

So that the potential is given by

(r) =q

4

1

r+ 1

r

=qd cos

4r2

Unlike an isolated point charge, the potential drops like1/r2 rather than 1/r



5/11

Electric Field of a Dipole

The potential calculation was easy since we didnt haveto deal with vectors

The electric field is simply E =

Since our answer is in spherical coordinates, but thereis no variation due to symmetry, we have

= rr

+ 1r

= r2qd cos 4r3

+ 1r

qd4r2

sin

Thus the electric field is given by

E = qd4r3

r2cos + sin



6/11

Electric Flux Density

It shall be convenient to define a new vector D in termsof E

D = E

Note that the units are simply C/m2.

We call this the flux density because the amount of itsflux crossing any sphere surrounding a point source isthe same

S

D

dS =

2

0

0

q

4r2

r2 sin dd ==q

4

22 = q



7/11

Gauss Theorem

Because of the 1/r2 dependence of the field, theintegral is a constant

Gauss law proves that for anysurface (not just asphere), the result is identical

S

D

dS = q

Furthermore by superposition, the result applies to anydistribution of charge

S

D dS =V

dV = qinside



8/11

Gauss Law Proof

First of all, does it makes sense? For instance, if weconsider a region with no net charge, then the fluxdensity crossing the surface is zero. This means that

the flux lines entering the surface from the left equal theflux lines leaving the surface on the right

We can prove that the flux crossing an infinitesimal

surface of any shape is the same as the flux crossing aradial cone

Notice that if the surface is tilted relative to the radial

surface by an angle , its cross-sectional area is largerby a factor of 1/ cos

The flux is therefore a constant

d = D dS = DrdScos = Dr dS

cos cos = DrdS



9/11

Application of Gauss Law

For any problem with symmetry, its easy to calculatethe fields directly using Gauss Law

Consider a long (infnite) charged wire. If the chargedensity is C/m, then by symmetry the field is radial

Applying Gauss law to a small concentric cylindersurrounding the wire

D dS = Dr2rd

Since the charge inside the cylinder is simply d

Dr2rd = d

Dr =

2rUniversity of California, Berkeley EECS 117 Lecture 8 p. 9/


10/11

Spherical Cloud of Charge

Another easy problem is a cloud of charge Q. Since the

charge density is uniform = Q/V, and V = 43

a3

D dS =

43r3 = Q

ra

3 r < a

Q r > a

ButD dS = 4r2Dr

Dr = Qr4a3

r < aQ

4r2 r > a



11/11

Charge Sheet

Consider an infinite plane that is charged uniformly withsurface charge density s

By symmetry, the flux through the sides of a centeredcylinder intersecting with the plane is zero and equal atthe top and bottom

The flux crossing the top, for instance, is simply DdS,

where D only can have a y component by symmetry

The total flux is thus 2DdS. Applying Gauss Law

2DdS = sdS

The electric field is therefore

E = y s

20 y > 0s20 y < 0University of California, Berkeley EECS 117 Lecture 8 p. 11/

Lecture 8 Electrostatic Potential & Gauss Law

Documents

Transcript of Lecture 8 Electrostatic Potential & Gauss Law