Mod-1 Electric Field

•

THE ELECTRIC FIELD  1. Charge  1.1. Microscopic Charge Carrier  1.1.1.Eletron

 1.1.2 Proton  2. Coulombs law  3. The electric field  4. The superposition of electric forces  5.1 Example: Electric field of Point Charge Q.  5.2 Example: Electric Field of a line of charges

 5.3 Example: Electric Field of Dipole  5.4 Example : Electric Field of ring of charge 6. Shell Theorem Example: Electric Field of Sphere of charge 7. Field Lines 8. Electric Dipole in an Electric Field 1.

CHARGE Charge is the source and the object of action of an electromagnetic field. 1.1 Microscopic Charge Carrier: Charge carriers are charged particles and ions which can carry both positive and negtive

Charge. Charge carriers like electron, proton and their antiparticles have infinite lifetime. Charge of ion is due to the fact that electron shell of atom or molecule lacks one or several electrons or, on the contratry, has extra electrons. Numerical value of elementary charge is |e| = 1.6021892 X 10-19 Coulomb 1.1.1 Electron. An electron is the material carrier of an elementary negative charge. It is usually assumed that an electron is a structureless point particle. Mass of electron is me = 9.1 x 10-31 kg 1.1.2 Proton. A proton is the carrier of positive elementary charge. Unlike an electron, a proton is not considered as a point particle. The entire charge of proton is concentrated in a sphere of radius 10-15 m 2.

Coulomb’s Law

Charles Augustin Coulomb deduced the law of interaction between two charged particles in 1785 from his experimental obsevations. “If two point charged particles are separated by a distance then the force of attraction or repulsion between them is proportional to the product of the magnitude of charges and inversely proportional to the square of the distance between them, the force acting along the line joining the charges” Let two point charges having magnitude q1 and q2 F12 q1 q2 F21 are separated by a distance of d. The electrostatic force of attraction or repulsion F12 F21 between them has the magnitude q1

q2

F= k

q1 q2 d2

In which k is proportionality constant. If the particles repel each other, the force on each particle points away from the other particle. If the two particles attract each other then the force on each particle points toward the other particle. For historical reasons, the electrostatic constant k is usually written as 1/4π∈o . Then the Coulomb’s law becomes

F=

1 q1 q2 4π ∈o d 2

The constant in the above equation have the value K = 1/4π∈o = 8.99 x 10 9 N.m2/C2.

3.

The Electric Field

The presence of an electric charge produces a force on all other charges present. The electric force produces action-at-a-distance; the charged objects can influence each other without touching. Suppose two charges, q1 and q2, are initially at rest. Coulomb's law allows us to calculate the force exerted by charge q2 on charge q1 (see Figure 3.1). At a certain moment charge q2 is moved closer to charge q1. As a result we expect an increase of the force exerted by q2 on q1. However, this change can not occur instantaneously (no signal can propagate faster than the speed of light). The charges exert a force on one another by means of disturbances that they generate in the space surrounding them. These disturbances are called electric fields. Each electrically charged object generates an electric field which permeates the space around it, and exerts pushes or pulls whenever it comes in contact with other charged objects. The electric field E generated by a set of charges can be measured by putting a point charge q at a given position. The test charge will feel an electric force F. The electric field at the location of the point charge is defined as the force F divided by the charge q: q1

q1

Fc

q2

Fc

q2

Figure 3.1. Electric force between two electric charges.

Ε=

F q

(3)

The definition of the electric field shows that the electric field is a vector field: the electric field at each point has a magnitude and a direction. The direction of the electric field is the direction in which a positive charge placed at that position will move. In this chapter the calculation of the electric field generated by various charge distributions will be discussed.

4.

THE SUPERPOSITION OF ELECTRIC FORCES

When there are more than two point charges then the interaction among themselves follow the following convention: (i) The force of interaction between two point charges does not change in the presence of the other charges. (ii) The force exerted on a point charge by two or more charges is equal to the sum of the forces exerted by each point charge separately in the absence of the other. q1

F23 q3

q2

F13

In the following figure the force F3 on the charge q3 due to q1 and q2 equal to the sum of the forces F13 and F23 .i.e. F3 = F13 + F23 (4.1) The above expression can be written as F3 = q3 E13 + q3 E23 (4.2) Where, E13 and E23 are the electric field strengths due to charges q1 and q2 at the position of q3. If E3 is the electric field at the position of q3 then F3 = q3 E3 (4.3) Comparing equation (4.2 ) and (4.3) we get E3 = E13 + E23 (4.4) This expression is the field form of the superposition principal.

5.1 EXAMPLE: ELECTRIC FIELD OF POINT CHARGE Q. A test charge placed a distance r from point charge Q will experience an electric forceFc given by Coulomb's law:

Fc =

1 qQ  r 4πε o r 2

(5.1.1)

The electric field generated by the point charge Q can be calculated by substituting eq.(5.1.1) into eq.(3)

 F 1 Q  E = c = r q 4πε o r 2

(5.1.2)

5.2 EXAMPLE: ELECTRIC FIELD DUE TO A LINE OF CHARGE A total amount of charge Q is uniformly distributed along a thin, straight, plastic rod of length L (Fig 5.1).

a) Find the electric field at a point P, at a distance d from one end of the rod ( Figure 5.2). b) Find the electric field at a point P', at a distance y from the midpoint of the rod ( Figure 5.3).

Figure 5.1 Problem 5.2 a) Electric field at a point P, at a distance d from one end of the rod Figure 5.2 shows the field at point P, as due to charge dq on a small segment dx of the rod. The force is directed along the x-axis and has a magnitude given by

Q dx 1 L dE = 4π ∈o ( d − x ) 2

(5.1)

Figure 5.2 Relevant dimensions for problem 5.2a. The resultant electric field at point P can be found by summing over all segments of the rod: −2 +1 0   1  Q dx Q Q  1  ( d − x ) E = ∫ dE = =     = 2 ∫ 4π ∈o L − L ( d − x ) 4π ∈o L  − 1   − 2 + 1  − L 4π ∈o L  ( d − x )  − L 0

0

1  1  d − d + L  Q L = 4π ∈o L d ( d + L ) =

Q 4π ∈o L

=

1 Q .... ..... (5.2) 4π ∈o d ( d + L )

b) Electric field at a point P', at a distance y from the midpoint of the rod Figure 5.3 shows a charged plastic rod lying along the x-axis with its center at the origin. A point P is situated at a distance of y from the center on the y-axis. Let us choose two charge segments equidistant from the center situated on both side. The electric fields at P due to these two charged segments are dEl and dEr respepctively. The net field dE at this point by the two segments of the rod is directed along the y-axis (vertical axis), and has a magnitude equal to dE = dEl cosθ + dEr cosθ or,

dE = dE l

y x2 + y2

+ dE r

y x2 + y2

(5.3)

[Note: the x-component of dEl cancels the x-component of dEr, and the net field is therefore equal to the sum of the y-components of dEl and dEr. The magnitude of dEl and dEr can be obtained from Coulomb's law:]

dE l = dE r =

1 4πε o

(

Q dx L x2 + y2

)

dE dEr

x = -L/2

dEl

x = L/2 dx

x x=0 Figure 5.3. Relevant dimension for problem 5.2b Substituting eq. (5.4) into eq. (5.3) we obtain

(5.4)

Q dx 1 L dE = 4πε o x 2 + y 2

(

)

3

(5.5)

2y 2

The net electric field at P can be obtained by summing over all segments of the rod.

1 Q E = ∫ dE = 2y 4πε o L

L/2

∫ (x 0

dx 2

+ y2

)

3

(5.6)

2

x = y tan θ

from the figure we can write

dx = y sec2 θ dθ

so that

and when x = 0 , θ = 0 and when x = L/2, θ = θ′ The integral in the above equation then becomes

I=

θ′

∫ 0

θ′

y sec 2 θ dθ 1 1 1 1 θ′ = 2 ∫ cosθ dθ = 2 [ sin θ ] 0 = 2 sin θ ′ = 2 3 3 y sec θ y 0 y y y

L

2 2 2 y +L

= 4

L 2 2y2 y2 + L

4

Substituting the value of above integral in the previous equation F the value of E is readily obtained as

E=

1 4πε o

Q L2 y + y2 4

(5.7)

If the length of charged rod is very large i.e. L>>y then y 2 can be neglected in the denominator of the above equation. And the expression becomes

E=

F 1 = q 4πε o

Q 2

y

L 4

=

1 Q 1 λ = 2πε o yL 2πε o y

(5.8)

Where λ = Q/L is linear charge density on the rod. 5.3 EXAMPLE: ELECTRIC FIELD DUE TO A DIPOLE An electric dipole is a configuration consisting of two-point charges equal in magnitude but opposite in sign separated by a distance. Figure below shows an electric dipole formed by two equal opposite charges of magnitude q and separated by a distance d. (a) Find the electric field due to the dipole at a point P, a distance z from the mid point of the dipole and on its axis. (b) Find the electric field due to the dipole at a point P, a distance z P from the mid point of the dipole and on the perpendicular bisector of its axis r(a) Figure 5.4a shows the electric fields due to charges +q and –q at the point P situated on the dipole axis. r+ z The electric fields E+ and E- due to the separate charges lie along the dipole axis that we take to be the z-axis. Applying superposition principal we +q find that the magnitude of E of the electric field at point P is d -q

Figure : 5.4 Relevant dimension for problem 5.4a

E = E+ − E− = =

1 q 1 q − 2 4π ∈o r+ 4π ∈o r−2 q d  4π ∈o  z −  2 

2

−

q d  4π ∈o  z +  2 

2

The above equation can be rewritten as −2 −2  q d  d    E=  − 1 +   1 − 2z  2 z   4π ∈o z 2  

Usually electrical effects due to a dipole are determined at distances that are large compared to the dimensions of the dipole where z>>d. At such distances we get d/2z <<1. Now using binomial theorem expanding the two quantities in the brackets in the above equation we get

    2d 2d + ....... − 1 − + ........ 1 + 2 z (1!) 2 z (1!)     So

 q d d    1 + + ....... − 1 − + ........ 2  4π ∈o z  z z    1 2qd = 2 4π ∈o z z

E =

=

1 qd 1 qd = 2 2π ∈o z z 2π ∈o z 3

.....

.......

......

(5.9)

The product qd is called electric dipole moment of the dipole and is denoted by p. p is a vector quantity whose direction is taken to be from negative end to the positive end of the dipole. Thus the electric field due to a dipole at an axial point is

E=

1 p 2π ∈o z 3

5.4 EXAMPLE: ELECTRIC FIELD DUE TO A RING OF CHARGE Q amount of charge is uniformly distributed over a insulating thin ring of radius R. the charge per unit length of the ring is

λ =

Q 2π R

Let us choose an infinitesimal length element dl at the top of the ring. The charge on that element is

dq = λ dl = dl

Q dl 2π R dE sinθ

R

P dE cosθ dE cosθ dE sinθ

The electric field dE due to this charge element at a point P on the axis of the ring is

dE =

1 dq 1 = 2 4πε o r 4πε o

 Q  dl   2 2  2π R  R + z

As dE makes an angle θ with the axis of the ring, it can be resolved into two mutually perpendicular components dE cosθ and dE sinθ respectively. It can be seen form the figure that there can be found a diametrically opposite element of the ring for which the electric field at P is dE’ which also makes an angle of θ with the axis. dE’ is equal in magnitude to dE and hence its perpendicular component dE’ sinθ cancels dE sinθ. Similarly the perpendicular components due to all the elements of the ring are cancelled out and hence the net electric field at P is determined by the summation of the components parallel to the axis ie dE cosθ. Here

cos sθ =

z R2 + z2

So that

E = ∫ dE cosθ =

1 Q dl 2 ∫ 4πε o 2πR R + z 2

1 Q 4πε o 2πR R 2 1 Q = 4πε o 2πR R 2 =

( (

=

1 Qz 4πε o R 2 + z 2

(

z + z2 z +z

)

3

2

z R2 + z2

dl ) ∫ 3

)

3

2

2

( 2πR ) .....

......

..... (5.10)

2

Now for small distances for which z<
E=

1 Qz =Kz 4πε o R 3

where k is a constant i.e. K =

1 Q 4πε o R 3

Again for distant points where R>>zthe equation 5.10 can be written as

E=

1 Qz 1 Q = 3 4πε o z 4πε o z 2

6. FIELD LINES The electric field can be represented graphically by field lines. These lines are drawn in such a way that, at a given point, the tangent of the line has the direction of the electric field at that point. The density of lines is proportional to the magnitude of the electric field. Each field line starts on a positive point charge and ends on a negative point charge. Since the density of field lines is proportional to the strength of the electric field, the number of lines emerging from a positive charge must also be proportional to the charge. An example of field lines generated by a charge distributions is shown in Figure 6

Figure 6. Electric field produced by two point charges q = + 4

7. ELECTRIC DIPOLE IN AN ELECTRIC FIELD The net force acting on a neutral object placed in a uniform electric field is zero. However, the electric field can produce a net torque if the positive and negative charges are concentrated at different locations on the object. An example is shown in Figure 5. The figure shows a charge Q located on one end of a rod of length L and a charge - Q located on the opposite end of the rod. The forces acting on the two charges are given by F+ = EQ z (7.1) F- = - EQ z (7.2)

Figure 7. Electric Dipole in an Electric Field. Clearly, the net force acting on the system is equal to zero. The torque of the two forces with respect to the center of the rod is given by

τ =

∑r × F

=

1 1 L QE sin θ + LQE sin θ 2 2

= L QE sin θ (7.1) As a result of this torque the rod will rotate around its center. If [θ] = 0 deg. (rod aligned with the field) the torque will be zero. The distribution of the charge in a body can be characterized by a parameter called the dipole moment p. The dipole moment of the rod shown in Figure 7 is defined as

P = LQ In general, the dipole moment is a vector which is directed from the negative charge towards the positive charge. Using the definition of the dipole moment from eq.7.1 the torque of an object in an electric field is given by τ = p xE

Summery

1 q1 q2 4π ∈o d 2 1 qQ  Fc = r Electric force on a charge q due to Q charge 4πε o r 2  F 1 Q  E = c = r The electric field generated by the point charge Q q 4πε o r 2 Electric force between two charge

F=

Electric field at point 3 due to two charges

E3 = E13 + E23

Electric field due to a line of charge at a point on the perpendicular bi sector

E=

1 4πε o

Q

L2 y + y2 4 Q 1 λ = yL 2πε o y

1 2πε o 1 Q Electric field due to a ring of charge at a point on the axis E = 2 4πε o R + z 2 Electric field due to a large line of charge at a distance of y

E=

Solved Problems Sample Problem 01 : Under what circumstances, if any, is the gravitational attraction between two protons equal to their electrical repulsion? Solution: Since the proton mass is 1.67 × 10 −27 Kg , the gravitational force between two protons that are a distance r apart is

Fgrav

(

Gm1m2 6.67 × 10 −11 N − m 2 Kg 2 × 1.67 × 10 −27 Kg = = r2 r2

The electric force between the protons is

)

2

=

1.86 × 10 − 64 N r2

Felec =

(

kq1q2 9 × 10 9 N − m 2 C 2 × 1.6 × 10 −19 C = r2 r2

)

2

=

2.3 ×10 −28 N r2

As every separation r, the electric force between the protons is greater than the gravitational force between them by a factor of more than 10 36 ; the forces are never equal. Sample Problem 02 : A test charge of + 1×10 −6 C is placed halfway between a charge of + 5 × 10 −6 C and a charge of + 3 × 10 −6 C that are 20 cm apart (Fig. X). Find the magnitude and direction of the force on the test charge.

q

q1

F2

F1

10 cm

q2

10 cm [Fig: X]

Solution: The force exerted on the test charge q by the charge q1 is 9 N .m 2 −6 −6 2 × 1 × 10 C × 5 × 10 C kq1q2 9 × 10 C F1 = = = +4.5 N 2 ( 0.1 m ) 2 r1

This force is taken to be positive because it acts to the right. The force exerted on the test charge q by the charge q2 is

kqq F2 = 2 2 = r2

9 × 10 9 N .m

2

C2

×1× 10 −6 C × 3 ×10 −6 C

( 0.1 m ) 2

= −2.7 N

This force is taken to be negative because it acts to the left. The net force on the test charge q is

F = F1 + F2 = +4.5 N − 2.7 N = +1.8 N

and it acts to the right, that is, toward the + 3 ×10 −6 C charge. Sample Problem 03 : The electric field in a certain neon sign is 5000 V/m. (a) What is the force this field exerts on a neon ion of mass 3.3 × 10 −26 kg and charge +e? (b) What is the acceleration of the ion? Solution (a): The force on the neon ion is

F = qE = eE = 1.6 ×10 −27 C × 5 ×10 3V / m = 8 ×10 −16 N

Solution (b): According to the second law of motion F = ma , and so here

a=

F 8 × 10 −16 N = = 2.4 × 1010 m 2 s m 3.3 × 10 −26 Kg

Sample Problem 04 : How strong an electric field is required to exert a force on a proton equal to its weight at sea level? Solution: The electric force on the proton is F = eE and its weight is mg . Hence eE = mg and −27 m mg 1.67 × 10 kg × 9.8 s 2 E= = = 1.02 × 10 −7 V −19 m e 1.6 × 10 C

Sample Problem 05 : A neutral water molecule

( H 2O ) in

its vapor state has an electric dipole moment of

−30

magnitude 6.2 × 10 C.m How far apart are the molecule’s centers of positive and negative charge? Solution: The Key Idea here is that a molecule’s dipole moment depends on the magnitude q of the molecules positive or negative charge and the charge separation d. There are 10 electrons and 10 protons in a neutral water molecule, so the magnitude of its dipole moment is p = qd = (10e )( d ) , In which d is the separation we are seeking and e is the elementary charge. Thus,

d=

p 6.2 ×10 −30 C.m = 10e (10 ) 1.6 ×10 −19 C = 3.9 × 10 −12 = 3.9 pm

(

)

Problem Sheet 1.

Four charges are arranged in a square with sides of length 2.5 cm. The two charges in the top right and bottom left corners are +3.0 x 10-6 C. The charges in the other two corners are -3.0 x 10-6 C. What is the net force exerted on the charge in the top right corner by the other three charges?

2.

Two identical spheres of mass m are hung from silk threads of length L, as shown in figure below. Each sphere has the same charge, so q1= q2 =q. Te radius of each sphere is small in comparison to the distance between them, so they may be treated as point charges. Show that if the angle θ is small, the equilibrium separation d between the spheres is

 q2L   d =   2πε o mg  3

1

θ L

3

m +qq

In the figure below three point charges are arranged at the corners of an equilateral triangle. Sketch the field due to +Q and – Q and from them determine the direction of the force that acts on because of the presence of these two charges.

5 6. 7.

+q

+Q

What is the magnitude of point charge that would create an electric field of 10 N/C at points 1.0m away? Two opposite charges of equal magnitude 2.0 x 10-7 C are held 15 cm apart. What are the magnitude and direction of E at the point midway between the charges? Plutonium 239 has a nuclear radius of 6.64 fm and the atomic number Z=94. Assuming that the positive charge is distributed uniformly what are the magnitude and direction of electric field at the surface of the charge In the figure below point charges +1.0 q and –2.0 q are fixed at a distant apart (a) find E at points A, B and C. d

A 8.

d/2

1.0 q

d/2

B

d

–2.0 q

+q

C

Figure below shows three charges at the corners of a isosceles right angled triangle. Calculate the magnitude and direction of the electric field a point P in the figure below.

a

P

+2q

9.

m q

d

+Q 4

L

a

+q

Q

In the figure below charges are placed on the vertices of an equilateral triangle. For what values of Q (both sign and magnitude) does the total electric field vanishes at C.

• +q

+q

10.

In the figure below four charges form the corners of a square and +3q four more charges lie at the midpoints of the sides of the square. The distance between adjacent charges on the perimeter of the a square is d. What are the magnitude and direction fo he electric field +q at the center of the square?

+5q

11.

12.

Four charges are arranged in a square with sides of length 5 cm. +q What are the magnitude and the direction of electric field at the center -8 square in figure if q = 1.0 x 10 C a +q Find the magnitude and direction of the electric field due to a electric dipole at a distant point situated on the d perpendicular bisector of its axis.

-q

+q

-5q

a

a -2q

+q a

+3q -2q

of

a aP

+2q

-q 14

An electron is constrained to the central axis of a ring of charge of radius R. show that the electrostatic force exerted on the electron can cause it to oscillate with an angular frequency

ω=

eq 4π ∈0 mR 3