ELECTRICITY & MAGNETISM (Fall 2011) LECTURE # 4 BY MOEEN GHIYAS.

ELECTRICITY & MAGNETISM (Fall 2011)

LECTURE # 4

BY

MOEEN GHIYAS

TODAY’S LESSON

(Electric Field & Gauss Law)

Fundamentals of Physics by Halliday / Resnick / Walker (Ch 23 / 24)

Today’s Lesson Contents

• Electric Field lines

• Motion of Charged Particles in Uniform Electric Field

• Electric Flux

Quiz # 1

• Time : 15 min

Electric Field Lines

• A convenient way of visualizing electric field patterns

is to draw lines that follow the same direction as the

electric field vector at any point. These lines, called

electric field lines, are related to the electric field in

any region of space in the following manner:

– The electric field vector E is tangent to the electric field line at

each point.

– The number of lines per unit area through a surface

perpendicular to the lines is proportional to the magnitude of

the electric field in that region. Thus, E is great when the field

lines are close together and small when they are far apart.


• Figure – Non uniform Electric Field

– Electric field is more intense on surface A than on surface B

– The lines at different locations point in different directions

indicates that the field is non-uniform.


• Figure - Electric field lines for the field due to a single charge

– The lines are actually directed radially from the charge in all

directions; thus, instead of the flat “wheel” of lines shown in

2 dimension, you should picture an entire sphere of lines



– Why lines are directed outward for +ve charge and inward for -ve?

– Because a positive test charge placed in this field would be

repelled radially away from the positive point charge while same

would be attracted toward the negative charge



– In either case of +ve or –ve charge, the lines are along the radial

direction and extend all the way to infinity.

– Note that the lines become closer together as they approach the

charge; this indicates that the strength of the field increases as

we move toward the source charge.


• The rules for drawing electric field lines are:

– The lines must begin on a positive charge and terminate on a

negative charge.

– The number of lines drawn leaving a positive charge or

approaching a negative charge is proportional to the magnitude

of the charge.

– No two field lines can cross.


• Is this visualization of the electric field in terms of field

lines consistent with Coulomb’s law?

• To answer this question, consider an imaginary spherical

surface of radius r concentric with a point charge.

• From symmetry, we see that the magnitude of the electric

field is the same everywhere on the surface of the

sphere.

• The number of lines N that emerge from the charge is

equal to the number that penetrate the spherical surface.


• The number of lines N that emerge from the charge is equal to

the number that penetrate the spherical surface.

• Hence, the number of lines per unit area on the sphere is

N/4πr 2 (where the surface area of the sphere is 4πr 2).

• Because 4π is constant, E is proportional to the number of

lines per unit area, i.e., E varies as 1/r 2; this finding is

consistent with Coulomb’s law equation.

• Also for two charges, if object 1 has charge Q1 and object 2

has charge Q2, then ratio of number of lines is N2 / N1 = Q2 / Q1


• Problems with model of qualitatively describing the

electric field with the lines?

1. We draw a finite number of lines from (or to) each

charge, which appear as if the field acts only in certain

directions at certain places; instead, the field is

continuous—that is, it exists at every point.

2. Danger of gaining the wrong impression from a two-

dimensional drawing of field lines being used to describe

a three-dimensional situation.


• The lines are nearly radial at points close to either

charge,

• The same number of lines emerge from each charge

because the charges are equal in magnitude.

• At great distances from the charges, the field is

approximately equal to that of a single point charge of

magnitude 2q (q+q = 2q).


• Now we sketch the electric field lines associated with

a positive charge +2q and a negative charge -q.

• In this case, the number of lines leaving +2q is twice

the number terminating at -q.

• Hence, only half of the lines that leave the positive

charge reach the negative charge.

• The remaining half terminate on a negative charge

we assume to be at infinity.

• At distances that are much greater than the charge

separation, the electric field lines are equivalent to

those of a single charge q (i.e. 2q-q = q).

Motion of Charged Particles in Uniform Electric Field

• When a particle of charge q and mass m is placed in an

electric field E, We know E = F / q

• Then electric force on the charge is , F = E q

• But, from Newton’s 2nd law we know F = m a

• Thus, we have Fe = E q = m a

• The acceleration of the particle is then a = E q / m

• If E is uniform (i.e. constant in magnitude and direction),

then the acceleration ‘a’ is constant.

• If q is +ve, then its acceleration is in the direction of the

electric field, and opposite for negatively charged particle


• The electric field in the

region between two

oppositely charged flat

metallic plates is

approximately uniform

• Suppose an electron of

charge -e is projected

horizontally into this field

with an initial velocity vi i.


• Because the electric field E in figure is

in the +y direction, the acceleration of

the electron is in the -y direction i.e.

• Since acceleration is constant, we can

apply the equations of kinematics in

two dimensions with vxi = vi and vyi = 0.

• After the time t, the components of its

velocity are


• Its coordinates after time t are

• Substituting the value t = x / vi from

eqn 1 into eqn 2, we see that y is

proportional to x2. Hence, the trajectory

is a parabola.

• After the electron leaves the field, it

continues to move in a straight line in

the direction of v, obeying Newton’s

first law, with a speed v > vi


• Since we are dealing with atomic

particles, note that we have neglected

the gravitational force acting on the

electron.

• For an electric field of 104 N/C, the

ratio of the magnitude of the electric

force eE to the magnitude of the

gravitational force mg is of the order of

1014 for an electron and of the order of

1011 for a proton.


• Example – An Accelerated Electron – An electron enters the

region of a uniform electric field as shown in fig, with v i =

3.00 x 106 m/s and E = 200 N/C. The horizontal length of the

plates is l = 0.100 m.

(a) Find the acceleration of the electron while it is in the

electric field.

Solution: e = 1.60 x 10-19 C,

m = 9.11 x 10-31 kg


• Example – An Accelerated Electron – An electron enters the

region of a uniform electric field as shown in fig, with v i =

3.00 x 106 m/s and E = 200 N/C. The horizontal length of the

plates is l = 0.100 m.

(b) Find the time it takes the electron to travel through the

field.

Solution:

Electric Flux – (Uniform Electric Field)

• Electric field lines - qualitative method

• Electric flux - quantitative way

• Consider an electric field that is uniform in

both magnitude and direction as shown

• The total number of lines penetrating the

surface is proportional to the product EA

• Product of the magnitude of the electric field

E and surface area A perpendicular to the

field is called the electric flux ΦE : ΦE = EA

• From SI units flux is measured as Nm2/C

Electric Flux• Example – What is the electric flux through a sphere

that has a radius of 1.00 m and carries a charge of

+1.00 μC at its centre?

• Solution:

• Area of sphere = 4πr2 = 12.6 m2

• Note: The field points radially outward and is therefore

everywhere perpendicular to the surface

Electric Flux

• What if the surface through which electric field passing

is not perpendicular to the field

Electric Flux

• Flux through a surface of fixed area A has

– A max value EA when the surface is perpendicular to the field

(when the normal to the surface is parallel to the field (θ = 00);

– the flux is zero when the surface is parallel to the field (when

the normal to the surface is perpendicular to the field (θ = 900);

Electric Flux – (Non Uniform Electric Field)

• If the electric field varies (non-uniform) over a given

surface and generally it does. Then, our definition of flux

is applied only over a small element of area.

• Consider a general surface divided up into a large

number of small elements, each of area ΔA. The

variation in the electric field over one element can be

neglected if the element is sufficiently small.

Electric Flux

• For vector ΔAi magnitude is the area of the ith element of

the surface and direction is perpendicular to the surface

element, as shown. The electric flux ΔΦE through this

element is

• By summing the contributions of all elements, we obtain

the total flux through the surface.

Electric Flux

• If we let the area of each element approach zero, then

the general definition of electric flux is

• Above equation is a surface integral, which means it

must be evaluated over the surface in question.

Electric Flux

• We are often interested in evaluating the flux through a

closed surface, that divides space into an inside and an

outside region

• Consider a closed

surface as in figure

Electric Flux• The net flux through the surface is proportional to the

net number of lines leaving the surface, where the net

number means the number leaving the surface minus

the number entering the surface.

• If more lines are leaving than entering, the net flux is

positive.

• If more lines are entering than leaving, the net flux is

negative.

Electric Flux• Using the symbol to represent an integral over a

closed surface, we can write the net flux ΦE as

• where En represents the component of the electric field

normal to the surface

Electric Flux• Example – Consider a uniform electric field E oriented

in the x direction. Find the net electric flux through the

surface of a cube of edges ℓ , oriented as shown.

Electric Flux• Solution: The net flux is the sum of the fluxes through

all faces of the cube.

• First, note that flux through four of the faces ( , , and

the unnumbered ones) is zero because E is perpendicular

to dA on these faces.

• Net flux through faces and is

33 44

11 22

Summary / Conclusion

• Electric Field Lines

• Motion of Charged Particles in Uniform Electric Field

• Electric Flux

ELECTRICITY & MAGNETISM (Fall 2011) LECTURE # 4 BY MOEEN GHIYAS.

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