Mod-2 Gauss Law

CHAPTER 2 1.

Introduction

2.

Concept of Flux

3.

Flux of Electric field

4.

Gauss's Law

5.

Some consequences of Gauss's applied to conductors

6.

Use of Gauss's Law to find the E-field due to symmetrical charge distributions (a) Point charge. (b) E due to charged conducting sphere (c) E due to charged non-conducting sphere (d) E due to non-conducting plane of charge (large) (e) E due to charged conducting plane

(f) E between two conducting plate with equal and opposite charge (g) E due to a line of charge

1.

Introduction

The electric field of a given charge distribution can in principle be calculated using Coulomb's law as we have done in Chapter1. However the actual calculations can become quite complicated in many cases. Gauss's law (which is deduced from Coulomb's law) provides a convenient technique for determining the E due to number different, high-symmetry charge distributions. Furthermore by suitable vector manipulation we arrive at the first of Maxwell's equations.

2.

Concept of Flux

If an electric fieldE passes through a surface A which is described by the vectorA, then the electric flux [ΦE] throughA is defined as the product of the area ofA and the magnitude of the normal component of the electric fieldE:

ΦE = A E cosθ = EA

…….

……

(2.1)

For a closed surface an outward directed flux is defined to have a positive sign whereas an inward flux has a negative sign

E

θ

A

If the E is not constant over the whole surface A then A must be divided into a series of elements dA over which E is constant. The total flux through A is then given by

φE =

∫ E •d A

( For open surface)

…….

A ……

(2.2)

where the integral is a surface one over A. However the surface over which the integral is evaluated may be closed. In that case the integration is done over the closed surface. i.e.

φE =

∫ E •d A

( For

closed

surface )

…….

……

(2.3)

The SI unit of electric flux is N-m2/C

Example 2.1 : A cylinder is immersed in a uniform electric field as shown in the figure below calculate the flux through its (a) left surface , (b) right surface (c) Cylindrical surface

dA A

A

3.

Gauss's Law

Gauss’s Law relates electric field at points on a closed surface ( called Gaussian Surface ) to the net charge enclosed by that surface. " If the volume within an arbitrary closed mathematical surface holds a net electric charge Q, then the electric flux [ФE] though its surface is Q/εo " Gauss' law can be written in the following form:

ΦE = or

∫ E •d A

=

Q εo Q εo

…….

……

(3.1)

…….

……

(3.2)

The circle on the integral sign indicates that the integral is to be taken over a closed surface. From the principle of superposition any number of charges contained within A produce an individual flux through the surface that add together to give a total flux. Any charges lying outside of A contribute equal amounts of inward and outward flux and hence do not contribute to the total flux through A.

4.

Some consequences of Gauss's law as applied to conductors

A conductor contains at least some charges which are free to move within it. When initially placed in an external E-field the field will penetrate into the conductor and cause the free charges to move (a). However these can only move as far as the surface of the conductor (assuming it is of finite size). Charge collects at the surface and produces an E-field within the conductor which opposes the external field. Equilibrium is quickly reached where the displaced charges produce an internal field which exactly cancels the external one (and hence there is no further movement of charge) (b).

⇒ Within a conductor at equilibrium there can be no E-field. All points of the conductor must be at the same potential. When a solid conductor in (a) carries a net charge. Within conductor E=0 hence flux through Gaussian surface G is zero and hence net charge contained within G is also zero. ⇒ A solid conductor carries all its excess charge on the surface

. Hollow conductor in figure (b) must also carry any excess charge on its outer surface unless the hollow region contains a charge (+Q) (figure (c)) in which case the inner surface must carry an equal but opposite charge -Q. These requirements are necessary to give a zero flux through the Gaussian surface G. It can also be shown that in case (b) E=0 within the hollow region

5.

Use of Gauss's Law to find the E-field due to symmetrical charge distributions (a) E due to a point charge. The field generated by a point charge q is spherical symmetric, and its magnitude will depend only on the distance r from the point charge. The field only have a radial component (see Figure 24.2). Consider a spherical surface centered around the point charge q (see Figure 24.2). The direction of the electric field at any point on its surface is perpendicular to the surface and its magnitude is constant. This implies that the electric flux [ФE] through this surface is given by ФE =

∫ E • d A = ∫ E dA = E ∫ dA

∫ dA is equal to the area of

……

…..

(5.1)

+q

spherical Gaussian surface and is equal to

4πr2 so that the flux is

ФE = E(4πr2)

……

…..

(5.2)

……

…..

(5.3)

……

…..

(5.4)

Gaussian surface

Now from Gauss’s law we can write E(4πr2) = q/εo Hence E =

1 q 4πε o r 2

(b) E due to charged conducting sphere (i) E inside the sphere A solid conductor carries all its excess charge on the surface. Let us consider a conducting sphere of radius a carrying a charge Q on its surface. Let us consider a Gaussian surface (G) a sphere of radius r (
r a

0 ∫ E •d A = εo = 0

Gaussian surface

so that E =0 inside the sphere (r
(ii) E outside the sphere A conducting spheres of radius a carries a charge Q on its surface. Because of symmetry in charge distribution the resultant electric field must by spherically symmetric and hence can only have a radial component. Let us consider a Gaussian surface (G) a sphere of radius r ( rel="nofollow">a) placed concentric to the conductor. By symmetry the field E must be constant at all points on G and also normal to G. Flux through G ФE = field x surface area = E.(4πr2 ) Total charge enclosed by surface is Q = εo (E.4πr2) So that the electric field E=

1 Q ……… 4π ∈o r 2

……..

(5.5)

E

E=

1 Q 4π ε o a 2

If we plot E as a function of distance(r) from the center of the Charged conducting sphere the graph will be as follows

a

r

(c) E due to charged non-conducting sphere (Charge distributed uniformly throughout the volume of a sphere.) (i) Inside the sphere Let us consider that charge Q is uniformly distributed throughout a volume of sphere of radius R. To find electric field inside the sphere r
1 Q′ 4πε o r 2

r

Now the charge enclosed by the Gaussian surface is

R

Qr3 4 Q 4 3 ρ π r3 = πr = 3 Q’ = 4 3 R π R3 3 3

Q` Gaussian surface

Substituting this result in the expression for E we get E=

  Q  r 3  4 π ∈ R o  

……

……

……..(5.6)

The quantity in the parentheses is a constant, so, within the spherically symmetric charge distribution electric field E is directly proportional to r.

(ii) Outside the sphere For points outside the sphere r>R, the charge produces an electric field on the Gaussian surface as if the charge were a point charge located at the center and the electric field is E=

(d)

1 Q 4π ∈o r 2

…….

…...

E=

E

(5.7)

a

1 Q 4π ε o R 2

r

E due to non-conducting plane of charge (large).

An infinite (or very large) sheet carries a uniform charge density σ. By symmetry the resultant Efield must have a direction normal to the plane and must have the same value at all points a common distance from the plane. Let us consider a Gaussian surface as a cylinder of cross-sectional area A and height 2h. The cylinder extends equally on both side of the plane of charge. Flux is only non-zero through ends of cylinder.

h

If field at cylinder ends is E then total flux is ФE = 2EA. Charge enclosed is q = area x charge density = Aσ Hence from Gauss's law we get εo( 2EA) =Aσ

A

so that the electric field at point near the plane of charge is

E=

σ 2ε o

E

……..

……..

(5.8)

+ +h + + + + + ++ + + + + + + +++ ++

E

A

(e) E near a charged conducting plane The electric field on the surface of a charged conducting plane can be found by applying Gauss' law (see Figure below). Here the charge is uniformly distributed over two surface of the conductor. E-is normal to the plane and have the same value at all points a common distance from the plane. Let us consider a Gaussian surface as a cylinder of cross-sectional area A and height 2h. The cylinder extends equally on both side of the plane of the conductor. Flux is only non-zero through the ends of cylinder. If field at cylinder ends is E then total flux is ФE = 2EA.

+ + + + + + + + + +

A

Charge enclosed is q = 2(area x charge density) = 2Aσ Hence from Gauss's law we get

E

εo( 2EA) =2Aσ So that the electric field at point near the plane of charge is

E=

σ εo

……..

……..

(5.9)

+ + + + + + + + + +

+ + + + + + + + + +

+ + + + + + + + + + + h + + + + + + + + + + + + + + + + + ++ + + + + + + + + + ++ + + + +++ ++ ++

A E

+

++ +++ +++

++ +

+

Charge

(f) E between two conducting plate with equal and opposite charge

When two charged conducting plate are brought close together the excess charges are induced to their inner surfaces. This is a special situation in which an uniform the electric field is created within the plates and electric field out side the combination is zero. The surface charge density on the inner plates is now double to that of an isolated charged plate. The electric field E field within the plates can be determined by applying Gauss’s law to the surface as shown in the figure below. The flux is non-zero only through the front face and is equal to ФE = E A

+

Charge enclosed is q = (area x charge density) = Hence from Gauss's law we get εo( E A) =Aσ` So that the electric field at point in between the plates is

E=

σ′ …….. εo

……..

(5.10)

surface charge E=0

here σ′ = 2σ where σ is the density of an isolated charged conducting plate

Aσ`

+ + + + + + + + + + +

+ + + + +

+ + + + + + + + + + + + +

E-

E=0

A-

-

fig (a) perspective view

+ + + E=0

A-

+ + + + + +

-

E

E=0

(b) side view

g)E due to infinite line of charge. Charge is uniformly distributed over the length of an very large insulating rod (plastic ) rod. The charge per unit length of the rod is λ. The electric field must have a direction normal to the axis of the rod and must have the same value at all points a common distance from it. Let us take as a Gaussian surface a coaxial cylinder of radius r, cross-sectional area A and length h. Flux is only non-zero through the cylindrical surface. Here E is perpendicular to the surface and is

+ + + + + + + + + + + + + + ++ + + + + + + + + + + + + + ++ + + + + + + + + + + + + + +

parallel to the surface vector at all points. If electric field at the cylindrical surface is E the flux through it is EA where A = 2πrh is the area of the cylindrical surface. Flux through the end faces are zero as the field lines are parallel to the faces. Hence the net flux through the Gaussian cylinder is ΦE = E A = E (2πrh ) Now from Gauss's law we get εo ΦE = hλ or. or,

εo E (2πrh ) = hλ

Ε=

1 λ 2 πε o r

The above expression gives the value of E at a distance r from a uniformly charged long rod

Problem Gauss’s law Ex : 1 q. Ex : 2

Use Gauss's law to calculate electric field at a distance, r, from a point charge, A 10 µC point charge is located at one of the corners of a cube. If the cube is 10 cm I on a side, what is the flux through each of the cube sides.

Ex : 3 For the constant electric field E = 105 N/C I , find the flux through a planar surface of area 1m2 when the surface is a) on the x-y plane. b) on the y-z plane. c) on the z-x plane.

y dA dA

Ex : 4 Consider a closed triangular box in the presence of a constant , horizontal electric field of magnitude E = 7.8 x 104 N/C, as shown in figure. Calculate the electric flux through a) the vertical face labeled A´ b) the slanted surface labeled A 30cm c) the entire surface of the box

10cm

Ex : 5

x dA z

600

A cube with 1.4 m edges is oriented as shown in figure below in a region of electric field (a) E = 3.00y j , (b) E = -4.00 I + 3.00 y j . E in Newton per Coulomb and y in meters. In both the cases determine the charge enclosed by the cube?

Ex : 6 Use Gauss's law to calculate E a distance , a, from a infinite wire with linear charge density, λ. Ex : 7 Use Gauss's law to calculate E at a distance, a, from a non-conducting infinite sheet . The sheet has a thick ness, t, and charge density, ρ. Ex : 8 Use Gauss's law to calculate E at a distance, a, from a conducting sheet with surface charge density,σ. Ex : 9 Figure below shows cross sections through two large , parallel , non-conducting sheets with identical distributions of positive charge with surface charge density σ. What is E at points (a) above the sheets, (b) between them, and (c) below them?

+ + + + + ++ + + + + + + + + + + ++ + + + + +

Ex:10 Two large thin metal plates are parallel and close to each other as in figure below, but with negative plate on the left. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 7.0 x 10-22 C/m2 . . What are the magnitude and direction of the electric field E (a) to the left of the h g plates (b) to the right s of the plates (c) between the Plates

-

+ + + + + + +

11 Figure below shows a section through two long thin concentric cylinders of radii a and b with a
E =

1 λ 4πε 0 r

12. The nucleus of an atom of gold has a radius R = 6.2 × 10 −15 m and a positive charge q = Ze, where the atomic number Z of gold is 79. Plot the magnitude of electric field from the centre of the gold nucleus outward to a distance of about twice its radius. Assume that the nucleus is spherical with a uniform charge distribution. 13. Charge is distributed uniformly through out the volume of an infinitely long cylinder of radius R. (a) Show that at a distance r from the cylinder axis (for r < R),

E=

ρr , where ρ is the 2ε 0

volume charge density. (b) Write an expression for E when r rel="nofollow"> R. 14. A thin-walled metal sphere has a radius of 25 cm and a charge of 2 × 10 −7 C . Find E for a point (a) inside the sphere, (b) just outside the sphere, and (c) 3.0 m from the centre. 15. A point charge causes an electric flux of -750 N.m2/C to pass through a spherical Gaussian surface of 10.0 cm radius centered on the charge. (a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface? (b) What is the value of the point charge? 16. A solid non-conducting sphere of radius R has a non-uniform charge distribution of volume charge density ρ = ρ s r R , where ρ s is a constant and r is the distance from the center of the sphere. Show that (a) total charge on the sphere is Q = πρ s R 3 and (b) the electric field inside the sphere has a magnitude given by E =

1 Q 2 r 4πε o R 4