1

Gauss’s law and its applications Lect. (3)

2

• The product of the magnitude of the electric field E and surface area A perpendicular to the field is called the electric flux E

• the SI units of electric flux (N.m2/C.)• Electric flux is proportional to the

number of electric field• lines penetrating some surface.

Electric Flux

3

Electric Flux

• The amount of electric field passing through a surface area• The units of electric flux

are N-m2/C

AE AdE

4

Gauss’s Law

The total flux passing through a closed surface is proportional to the charge enclosed within that surface.

Note: The area vector points outward

0q

AdEsurface

E

5

The Gaussian Surface

An imaginary closed surface created to enable the application of Gauss’s Law

6

1. The Electric Field Due to a Point Charge

2. The Electric Field Due to a Thin Spherical Shell

3. A Cylindrically Symmetric Charge Distribution

Application of Gauss’s Law to Various

Charge Distributions

7

1. Spherically Symmetric Charge Distribution

• The area vector is parallel to the electric field vector at the surface of the spherical Gaussian surface.• Applying Gauss’ law:

33

3

22in

2in

2in

22

R

rQk

R

rQ

r

k

r

qk

rπ4

qkπ4E

rπ4Eqkπ4

rπ4EAEΦrπ4Asphere1θcos

θcosAEdAEdAEΦ

1.The Electric Field Due to a Point Charge:

8

• As r increases to R, the magnitude of the electric field will increase as more charge is included within the volume of the spherical Gaussian surface.• The maximum electric field will be obtained when r = R.• For a spherical Gaussian surface of radius r R, we can consider the charged insulating sphere as a point charge.

1. Spherically Symmetric Charge Distribution

9

• Conclusion: the electric field E inside a uniformly charged solid insulating sphere varies linearly with the radius until reaching the surface. The electric field E outside the sphere varies inversely with 1/r2. The diagram on the next slide illustrates this relationship.

22in

2in

2inin

22

r

Qk

r

qk

rπ4

qkπ4

rπ4Eqkπ4qkπ4Φ

rπ4EΦrπ4Asphere1θcos

θcosAEdAEdAEΦ

E

1. Spherically Symmetric Charge Distribution

10

Spherically Symmetric Charge Distribution

11

2. The Electric Field Due to a Thin Spherical Shell

A thin spherical shell of radius a has a total charge Q distributed uniformly over its surface . Find the electric field at points(A) outside and (B) inside the shell.

12

• Consider a thin spherical shell of radius R with a total charge Q distributed uniformly over its surface.

• For points outside the shell, r R: the spherical Gaussian surface will enclose the thin spherical shell, so qin = Q.

2. The Electric Field Due to a Thin Spherical Shell

13

• The area vector is parallel to the electric field vector at the surface of the spherical Gaussian surface. The angle between the area vector and the electric field vector is zero.

22in

2in

2inin

22

r

Qk

r

qk

rπ4

qkπ4

rπ4Eqkπ4qkπ4Φ

rπ4EΦrπ4Asphere1θcos

θcosAEdAEdAEΦ

E

2. The Electric Field Due to a Thin Spherical Shell

14

• For points inside the thin spherical shell, r < R:

• The charge lies on the surface of the thin spherical shell.

• No charge is found within the shell, so qin = 0 and the electric field and flux within the thin spherical shell is also 0.

2. The Electric Field Due to a Thin Spherical Shell

15

Cylindrically Symmetric Charge Distribution

• Consider a uniformly charged cylindrical rod of length L.

16

Cylindrically Symmetric Charge Distribution

17

Cylindrically Symmetric Charge Distribution

• Instead of the shortened Gaussian cylinder indicated in the diagrams, I use a Gaussian cylinder that is the same length as the charged cylindrical rod.• The electric field lines are perpendicular to the charged

cylindrical rod and are directed outward in all directions. • The electric field is perpendicular to the Gaussian

cylinder and is parallel to the area vector at those points where it passes through the Gaussian cylinder. The angle between the area vector and the electric field vector is zero.

18

Cylindrically Symmetric Charge Distribution

• The equation for the area of a cylinder is: A = 2··r·L• Applying Gauss’ law:

r

λk2

Lr

Qk2

Lr

qk2

Lrπ2

qkπ4E

Lrπ2Eqkπ4qkπ4Φ

Lrπ2EΦLrπ2Acylinder1θcos

θcosAEdAEdAEΦ

inin

inin