unit 3 - electrostatics

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UNIT 3:ElectrostaticsUNIT 3:Electrostatics

The study of electric The study of electric charges at rest, the charges at rest, the

forces between them forces between them and the electric fields and the electric fields associated with them.associated with them.

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3.1 Electric Charges and Conservation of charges

{ The electric charge has the following important properties :z There are two kinds of charges in nature positive and negative

charge.{ Charges of opposite sign attractopposite sign attract one another attractive

force.{ Charges of the same sign repelsame sign repel one another repulsive

force.zz The total charge in an isolated system is constant The total charge in an isolated system is constant

(conserved)(conserved) Principle of conservation of chargesPrinciple of conservation of chargesz Charge is quantized.quantized.

{ Electric charge exists as discrete packets and written as neq =

C10 x 6.1e 19- charge, ofamount lfundamenta :

charge electric :qwhere

1,2,... number integer : =n

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{ Electrical conductorsz Definition is defined as the materials in which some of the

electrons (negative charges) are free electrons that are not bound to atoms and can move relatively freely through the material.

z For example : copper, aluminum, silver and etc{ Electrical insulators

z Definition is defined as the materials in which all electrons are bound to atoms and cannot move freely through the material.

z For example : glass, rubber, wood and etc...{ Explanation of conductors and insulators

z Consider two metal spheres, one highly charged and the other electrically neutral (number of positive and negative charge are number of positive and negative charge are equalequal) as shown in figure (a).

z Figure (b) shows the two spheres connected by a metal nail, which conducts charge (electron) from one sphere to the other.

z Figure (c) shows the two spheres connected by a wood, almost no charge is conducted.

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3.2 Coulombs Law{ States the magnitude of the electrostatic (Coulomb/electric) the magnitude of the electrostatic (Coulomb/electric)

force between two point charges is proportional to the force between two point charges is proportional to the product of the charges and inversely proportional to the product of the charges and inversely proportional to the square of the distance between them.square of the distance between them.

Mathematically,

chargespoint obetween tw distance :r

221

rqkqF =

2-29 C N m10 x 0.9k constant (Coulomb) ticelectrosta : =

221

rqqF

whereforce (Coulomb) ticelectrosta of magnitude :F

charge of magnitude :, 21 qq

++ ++r

2q1qFr

Fr

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{ Since

{ If q1 and q2 are charges of opposite sign, the force (F) acting on each charge is attractive as shown in figure below.

z This mean that F is directed towards the neighbouring charge and will result in both charges moving towards each other.

{ If q1 and q2 are both positive or both negative charges, the force (F) acting on each charge is repulsive.

z This mean that F is directed away from the neighbouring charge and will result in a separation of the two charges if they are free to move.

041k = , hence the Coulombs law can be written as

= 2 21

0 rqq

41F

whereair)or (vacuum space free ofty permittivi :0

) .( 212120 mNC10x858=

++r

--2q1q F

rFr

Simulation

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{ Graphs below show the variation of electrostatic force with the distance between two charges.

{ The S.I. unit of charge is coulombcoulomb (C).{ 1 Coulomb is defined as the total charge transferred by a current the total charge transferred by a current

of one ampere in one second.of one ampere in one second.{ Note :

z The sign of the charge can be ignored when substituting into theCoulombs law equation.

z The sign of the charges is important in distinguishing the direction of the electric force .

r

F

0

F

2r1

0

21qkqM =Gradient,

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{ Example 1 :

Two point charges, q1=-20 nC and q2=90 nC, are separated by a distance of 4.0 cm as shown in figure below.

Find the magnitude and direction of a. the electric force that q1 exerts on q2.b. the electric force that q2 exerts on q1.(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)

Solution: q1=2.0 x 10-8 C, q2=9.0 x 10-8 C, r=4.0 x 10-2 m

21F12 chargeon chargeby force :r

--cm04 .

++ 2q1q

--cm04 .

++ 2q1q 12Fr

21Fr

where

12F21 chargeon chargeby force :r

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a. By applying the Coulombs law equation :

b. By using the Coulombs law equation :

Conclusion :z The magnitude of both forces is the same but opposite in direction

obey the Newtons third law.

z The characteristic of electric force exert on both charges is attractive force.

22

889

12 10x410x0910x0210x09F

)().)(.)(.(

=r

2112 FFrr =

221

12 rqkqF =r

N10x01F 212= .r Direction : Direction : to the left (qto the left (q11))

212

21 rqkqF =r

N10x01F 221= .r Direction : Direction : to the right (qto the right (q22))

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{ Example 2 :

Three point charges lie along the x-axis as shown in figure below.

Calculate the magnitude and direction of the total electric force exerted on q2.(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)

Solution: r12=3.0 x 10-2 m, r23=5.0 x 10-2 m

By applying the Coulombs law equation :

32Fr12F

r

C4q2 =C2q1 =--++ ++

C6q3 =cm03 . cm05 .

C4q2 =C2q1 =--++ ++

C6q3 =cm03 . cm05 .

212

2112 r

qkqF =r

N10x08F 1312 .=r

Direction : Direction : to the right (qto the right (q33))

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and

Therefore, the total force exerted on q2 is given by

{ Example 3 :

Figure below shows the three point charges are placed in the shape of triangular.

223

3232 r

qkqF =r

N10x68F 1332 .=r


32122 FFFrrr +=

N10x616F 132 .=r

32122 FFFrrr +=

13132 10x6810x08F .. +=r


2q1q++

3q

--

--

12r13r

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212

2121 r

qkqF =r

Determine the magnitude and direction of the resultant electric force exerted on q1. Given q1=-1.2 C, q2=+3.7 C, q3=-2.3 C, r12=15 cm, r13=10 cm, =32 and k = 9.0 x 109 N m2 C-2.Solution: q1=1.2x10-6 C, q2=3.7x10-6 C, q3=2.3x10-6 C,

r12=15x10-2 m, r13=10x10-2 m

z By applying the Coulombs law equation :

Magnitude of F21:

N781F21 .=r

31Fr

21Fr

o582q1q

++

3q

--

--

12r13r

22

669

21 10x1510x7310x2110x09F

)().)(.)(.(

=r

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Magnitude of F31:

{ Construct a table to represents x and y-component for all forces exerted on q1.

{ Vector sum the x-comp. and y-comp. :

{ The magnitude of resultant electric force exerted on q1 :

213

3131 r

qkqF =r

( ) ( )2y12x11 FFF +=r

N482F31 .=r

N783F1 .=r

y-component(N)x-component(N)Force

21Fr

21Fr

0

31Fr or 58F31 cos o

r58F31 sin

N09358FFF 3121x1 .cos =+= orrN10258F0F 31y1 .sin == or

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{ The direction of resultant electric force exerted on q1 :

{ Example 4 :Two identical point charges A and B, each of mass 20 g, suspended from a fixed point O on two insulating threads as shown in figure below.

=

x1

y1

FFtan

o234.= or 325.8or 325.8 from the +xfrom the +x--axis axis (anticlockwise).(anticlockwise).

The charges are in equilibrium and each carries the same amount of charge, q. If =20, calculate a. the magnitude of both point charges. b. the magnitude of the electric force acting

on each charge.

(Given 0=8.85 x 10-12 C2 N-1 m-2)A

Bcm015 .

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{ Since the charges are in equilibrium, thus the point charge A also in equilibrium then

{ By dividing eq. (1) with eq. (2),

because

then eq. (3) becomes

= 0Fx

Solution: mA=mB=m=20x10-3 g, r=15.0x10-2 m, =20a. The free body diagram of point charge A :

sinTFe =

tan=mgFe

Tr

gmr

sinT

cosT

eFr

and = 0Fy(1)(1)

cosTmg = (2)(2)

(3)(3)

and20

BAe r4

qqF = qqq BA ==

tan=mgr4q

20

2

C10x24q 7= .

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b. By applying Coulombs law equation :

{ Example 5 : (exercise)Two point charges are placed on the x-axis as follows :Charge q1 = +4.00 nC is located at x = 0.200 m, charge q2 = +5.00 nCis at x = -0.300 m. Find the magnitude and direction of the total electric force exerted by these two charges on a negative point charge q3 = -6.00 nC that is placed at the origin. (Young & freedman,pg.829,no.21.20)(Given 0=8.85 x 10-12 C2 N-1 m-2)Ans. : 2.4 N to the right

N0710Fe .=

20

BAe r4

qqF =

20

2

e r4qF =

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{ Example 6 : (exercise)

Four identical point charges (q = +10.0 C) are located on the corners of a rectangle as shown in figure below.

The dimension of the rectangle are l = 60.0 cm and w = 15.0 cm. Calculate the magnitude and direction of the resultant electric force exerted on the charge at the lower left corner by the other three charges. (Serway & Jewett, pg. 735, no. 57)(Given 0=8.85 x 10-12 C2 N-1 m-2)Ans. : 40.9 N at 263

q

l

++ ++

++++

w

q

qq

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3.2.1 Relation of electric force, Fe and centripetal force, Fc{ Consider an electron(-ve) orbiting the nucleus (+ve) of an atom in

circular orbit of radius, r at tangetial (linear) speed, v as shown in figure below.

{ Since qe=qp=e thus

ce FFrr =

rvm

rqkq 2e2

pe =

{ The electric force between electron and nucleus contribute the net force (centripetal force).

2e

2

vmr

ke =kg10x11.9m 31-e masselectron : =where

constant c)ectrostatiCoulomb(el :kC10x60.1e 19- chargeelectron : =

v

relectronelectron

nucleusnucleus

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CoulombCoulombs laws lawNewtonNewtons law of Gravitations law of Gravitation

3.2.2 Comparison between Newtons Law of Gravitation and Coulombs Law.

Only attractive force

221

e rqkqF =

221

g rmGmF =

Attractive or repulsive force

Force due to mass interaction Force due to charge interaction

The force is a long-range forces.

The force is a short-range force.

The equation of the gravitational force :

The equation of the electric force :

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3.3 Electric Field{ Definition is defined as a region of space around isolated charge a region of space around isolated charge

where an electric force is experienced if a positivewhere an electric force is experienced if a positivetest charge placed in the region.test charge placed in the region.

{ Electric field around charges can be represented by drawing a series of lines. These lines are called electric field lines electric field lines (lines of force).

{ The direction of electric field is tangent to the electric field line at each point.

{ Figures below show the electric field patterns around the charge.a. Single positive charge b. Single negative charge

(the lines point (the lines point radiallyradially outward outward from the charge)from the charge)

(the lines point (the lines point radiallyradially inward inward toward the charge)toward the charge)

+q+q--qq

Field directionField direction

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+q+q --qq

c. Two equal point charges of opposite sign, +q and -q

d. Two equal positive charges, +q and + q

+q+q +q+q

(the lines are curved (the lines are curved and they are directed and they are directed from the positive from the positive charge to the negative charge to the negative charge.charge.

(point (point XX is neutral point )is neutral point ){{ is defined as is defined as a point a point

(region) where the total (region) where the total electric force is zero.electric force is zero.

{{ It lies alongIt lies along the vertical the vertical dash line.dash line.


XXField directionField direction

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+2q+2q--qq

e. Two opposite unequal charges, +2q and -q

f. Two opposite charged parallel metal plates

(note that twice as many (note that twice as many lines leave +2q as there lines leave +2q as there are lines entering are lines entering q, q, number of lines is number of lines is proportional to proportional to magnitude of chargemagnitude of charge.).)

{{ The electric field lines are perpendicular to the The electric field lines are perpendicular to the surface of the metal plates.surface of the metal plates.

{{ The lines go directly from positive plate to the The lines go directly from positive plate to the negative plate.negative plate.

{{ The field lines are parallel and equally spaced in the The field lines are parallel and equally spaced in the central region far from the edges but fringe outward central region far from the edges but fringe outward near the edges. Thus, in the central region, the near the edges. Thus, in the central region, the electric field has the same magnitude at all points.electric field has the same magnitude at all points.

{{ The fringing of the field near the edges can be The fringing of the field near the edges can be ignored because the separation of the plates is small ignored because the separation of the plates is small compared to their size.compared to their size.


Simulation

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g. Two equal negative charges, -q and -q (exercise).

h. Two unequal negative charges, -2q and -q (exercise).

{ The properties of electric field lines:z The field lines indicate the direction of the electric field (the field the field

points in the direction tangent to the field line at any pointpoints in the direction tangent to the field line at any point).z The lines are drawn so that the magnitude of electric field is

proportional to the number of lines crossing unit area perpendicular to the lines. The closer the lines, the stronger the closer the lines, the stronger the fieldfield.

z Electric field lines start on positive chargesstart on positive charges and end on end on negative chargesnegative charges, and the number starting or ending is proportional to the magnitude of the charge.

z The field lines never crossnever cross because the electric field dont have two value at the same point.

--qq --qq

--qq--2q2q

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3.4 Electric Field Strength (intensity),{ The electric field strength at a point,

Definition is defined as the electric (electrostatic) force per unit the electric (electrostatic) force per unit positive charge that acts at that point in the same positive charge that acts at that point in the same direction as the force.direction as the force.

Mathematically,

{ It is a vectorvector quantity.{ The units of electric field strength is N CN C--11 or V mor V m--11.

{ Since

Er

Er

0qFE =

20

rkqqF =

force electric theof magnitude :F

where

charge test of magnitude :0q

strength field electric theof magnitude :E

, then the equation above can be written as

=

0

20

qr

kqq

E2r

kqE = or 20r4

qE =where

chargepoint isolated of magnitude :qchargepoint isolated andpoint ebetween th distance :r

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{ Note :

z The direction of the electric field strength, E depends on the sign of isolated point charge.

z The direction of the electric force, F depends on the sign of isolated point charge and test charge. For example{ A positive isolated point charge.

a. positive test charge

b. negative test charge

qq )( veq0 +ErFr

r

qq )( veq0 Er

Fr

r

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{ A negative isolated point charge.a. positive test charge

b. negative test charge

z In the calculation of magnitude E, substitute the magnitude of magnitude of the charge only.the charge only.

qq )( veq0 +Er

Fr

r

qq )( veq0 Er F

r

r

Simulation

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{ Example 7 :Two point charges, q1=1 C and q2=-4 C, are placed 2 cm and 3 cm from the point A respectively as shown in figure below.

Finda. the magnitude and direction of the electric field intensity at point A.b. the total electric force exerted on q0=-4 C if it is placed at point A.(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)

Solution: q1=1 C, q2=4 C, q0=4 C, r1=2x10-2 m, r2=3x10-2 m

a. By applying the equation of electric field strength, the magnitude of

E at point A.Due to q1 :

++ -- 2q1qcm2 cm3

A

++ -- 2q1qcm2 cm3

A 1AEr 2AEr

22

9

21

11A 10x2

110x09rkqE

)())(.(

==113

1A CN10x252E= . Direction : Direction : to the right (qto the right (q22))

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Due to q2 :

therefore the electric field strength at point A due to the charges is given by

b. From the definition of the electric field strength,

thus the total electric force exerted on q0 is given by

22

9

22

22A 10x3

410x09rkqE

)())(.(

==113

2A CN10x4E= Direction : Direction : to the right (qto the right (q22))

2A1AA EEErrr +=

0

AA q

FE =

1313A 10x410x252E += .

r

113A CN10x256E

= .r Direction : Direction : to the right (qto the right (q22))

A0A EqF =

N10x52F 14A .=).)(( 13A 10x2564F =

Direction : Direction : to the left (qto the left (q11))

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{ Example 8 : (exercise)Find the magnitude of the electric field at point P due to the four point charges as shown in the figure below if q=1 nC and d=1 cm.

{ Example 9 : (exercise)Find the magnitude and direction of the electric field at the centre of the square in figure below if q=1.0x10-8 C and a= 5cm.

(Given 0=8.85 x 10-12 C2 N-1 m-2)(HRW. pg. 540.11)

Ans. : zero.

(Given 0=8.85 x 10-12 C2 N-1 m-2)(HRW. pg. 540.13)

Ans. : 1.02x105 N C-1, upwards.

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{ Consider a stationary particle of charge q0 and mass m is placed in a uniform electric field E, the electric force Fe exerted on the charge is given by

{ Since only electric force exerted on the particle, thus this force contributes the net force, F and causes the particle to accelerate.

{ According to Newtons second law, then the magnitude of the acceleration of the particle is

{ Because the electric field is uniform (constant in magnitude anddirection) then the acceleration of the particle is constant.

{ If the particle has a positive charge, its acceleration is in the direction of the electric field (figure 3.5a). If the particle has a negative charge (electron) , its acceleration is in the direction opposite the electric field (figure 3.5b).

3.5 Motion of Charged Particles in a Uniform Electric Field

EqF 0e =

maFF e ==maEq0 =

mEqa 0=

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{ Consider an electron (e) with mass, me enters a uniform electric field, E perpendicularly with velocity v0, the downward electric force will cause the electron to move along a parabolic path towards lower plate (figure 3.5c).

Fig.3.5aFig.3.5a

Er

eFr

ar

Fig.3.5bFig.3.5b

Er

eFra

r

Er y

0v

v

x

0q

Fig.3.5cFig.3.5cSimulation

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z Therefore the acceleration of the electron is given by

The negative sign indicates the direction of the acceleration is in y axis (downward).

z From the figure 3.5c, the path is similar to the motion of a ball projected horizontally above the ground.

{ The component of its velocity at time t are given by.x-component :y-component :

{ The position of the electron at time t is

jmeEa

e

=r

constant== 0x vv0v y0 =tavv yy0y += and

tmeEv

ey =

tvx 0=ta

21tvy yy0 =

2

e

tmeE

21y

=

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{ Electric flux of a uniform electric fielduniform electric field,Definition is defined as the scalar product between the electric the scalar product between the electric

field strength, E with the vector of the surface arefield strength, E with the vector of the surface area, A.a, A.Mathematically,

{ It is a scalarscalar quantity.{ The unit of electric flux is N mN m22 CC--11.

{ Consider a uniform electric field E passing through a surface area A as shown in figures 3.6a and 3.6b.From the fig. 3.6a, =0, thus

3.6 Electric Flux, E

AEErr =

area surface of vector theof magnitude :A

or

A and Ebetween angle :rr

strength field electric theof magnitude :E

= cosEAEwhere

o0EAE cos=EAE =

Er

A r

Fig. 3.6aFig. 3.6a

area, A

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{ From the fig. 3.6b, the angle between E and A is , thus

{ Note that the direction of vector A is always perpendicularperpendicular(normal) to the surface area, A.

{ The electric flux is proportional to the number of field lines electric flux is proportional to the number of field lines passing through the area.passing through the area.

{ Let us consider the more general case, when the electric field E is not uniform and the surface is not flat as shown in figure 3.6c.

EAE cos=

area, A

Fig. 3.6bFig. 3.6b

Ar

Er

Fig. 3.6cFig. 3.6cEr

Er

iAr

{ We divide up the chosen surface into n small elements of surface whose areas are UA1, UA2, UAn.

{ We choose the division so that each of UAi is small enough that{ it can considered flat.

{ the electric field can be considered uniform over this tiny area.

SF027 34Fig. 3.6dFig. 3.6d

{ Then the electric flux through the entire surface is approximately

{ If and the relation becomes mathematically exact :

{ In many cases, we deal with the flux through a closed surface and the net flux through the surface is given by

where the integral sign is written to indicate that the integral is over the value of E on an enclosing surface.

{ Note :The direction of vector dA is always point outward from the enclosed surface as shown in figure 3.6d.

=

=n

1iiiE AErr

= AdEE rr

0Ai r

, the sum becomes an integral over entire surface and

= AdEE rr

or = cosEdAE

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rq

Adr

Gaussian Gaussian surfacesurface

Er

Fig. 3.7aFig. 3.7a

{ States The total electric flux through an enclosed surface is The total electric flux through an enclosed surface is proportional to the algebraic sum of the electric proportional to the algebraic sum of the electric

charges within the surface.charges within the surface.Mathematically,

3.7.1 Equivalent of Gausss law and Coulombs law

{ Figure 3.7a shows a positive point charge q, around which we have drawn a concentric spherical Gaussian surface of radius r.

3.7 Gausss Law

0

encE

qAdE == rr

area surface of vector theof magnitude :dAsurface enclosedin charges electric theof sum algebraic :encq

strength field electric theof magnitude :Ewhere

space free ofty permittivi :0

{ From the Gausss law :

0

encE

qAdE == rr

0

encE

qEdA == cosand o0=

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{ Since the electric field strength, E is considered uniform hence

{ Because then

{ Exercise: Derived GaussDerived Gausss law by using Coulombs law by using Coulombs law.s law.

0E

qdAE ==

20r4

qE =

20

rqkqF =

0

2E

qr4E == )((surface area of the Gaussian surface)

EqF 0=

2r4dA =0

encE

qEdA == and

and04

1k =

2rkqE =

CoulombCoulombs Laws Law

qqenc =and

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3.7.2 Electric field of a Uniformly Charged Insulating Sphere

{ Consider an insulating solid sphere of radius R has a uniform volume volume charge densitycharge density and carries a total positive charge q as shown in figure 3.7b and 3.7c.

{ To find the magnitude of electric field inside the sphere, a spherical gaussian surface smaller than the sphere is drawn (figure 3.7b).

z This gaussian surface having the radius r < R.z Denote the volume of this small sphere is V and the charge

inside is qin (qin < q)z The volume charge density of a insulating solid sphere is given

by

==

3R34

qVq

R

rGaussian surface

Fig. 3.7b

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

+ rR

Gaussian surfaceFig. 3.7c

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

+

constantconstant

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z Therefore the charge inside the smaller gaussian surface qin is

z The magnitude of the electric field magnitude of the electric field is constantconstant and normalnormal to the surfacesurface at each point everywhere on the gaussian surface both conditionsconditions.

z By applying Gausss law ,

then the magnitude of the electric field inside the sphere:

== 30

32

Rqr)r4E(dAE

3

33

3in R

qrr34

R34

qV'q =

==

0

inE

qEdA ==

rR4

qE 30

= Inside (r

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{ To find the magnitude of electric field outside the sphere, a spherical gaussian surface bigger than the sphere is drawn (figure 3.7c).

z This gaussian surface having the radius r > R.z Denote the charge outside is qout and qout = qz By applying Gausss law ,

then the magnitude of the electric field outside the sphere:

{ Hence the graph of E against r for a uniformly charged insulating sphere can be shown in figure 3.7d.

==0

2

q)r4E(dAE

0

outE

qEdA ==

20 r

14qE

= Outside (r>R)Outside (r>R)

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3.7.3 Electric field of a Charged Conducting Sphere{ Consider a solid conducting sphere of radius R carries a net (total)

positive charge q as shown in figure 3.7e.

rE

Fig. 3.7d

R+ + ++++++

+++ ++ ++

++

E

r0 R

20 R

14qE

=

2r1E

Gaussian surfaceGaussian surface

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

+++R

r

0E =r

Fig. 3.7e

{ From the figure 3.7e, the positive charge resides on its surface only

{ It is because the conducting sphere in electrostatic equilibrium (no net electrostatic equilibrium (no net motion of charge within a motion of charge within a conductor)conductor).

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{ Then the magnitude of the electric field inside the conducting sphere is given by

{ To find the magnitude of electric field outside the sphere, a spherical gaussian surface bigger than the sphere is drawn (figure 3.7e).z For this choice, both conditions are satisfied, as they were for the

insulating sphere in subtopic 3.7.2.

z This gaussian surface having the radius r > R and the charge outside is qout = q.

z From the Gausss law,

then the magnitude of the electric field outside the sphere:

0E =

0

outE

qEdA ==

20 r

14qE

= Outside (r>R)Outside (r>R)

Inside (r

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3.7.4 Electric field of a Line Charge.{ Consider an infinitely long charged thin wire where the electric charge

is distributed uniformly on the wire. The wire has a line of positive charge of infinite length and constant charge per unit length charge per unit length as shown in figure 3.7g.

{ The symmetry of the charge distribution requires that E be perpendicular to the line charge and directed outward as shown in figure 3.7h.

{ To find the magnitude of the electric field at distance r from the wire, a cylindrical gaussian surface is drawn (figure 3.7g).

Fig. 3.7gl

Er

Adr

r

0E =Adr

Er

Er

wire

Fig. 3.7h

Gaussian surface

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z The gaussian surface having the length l and radius r.z For this choice, both conditions are satisfied, as they were for the

insulating sphere in subtopic 3.7.2.z The charge inside the cylindrical gaussian surface is given by

z By using the Gausss law,

then the magnitude of the electric field at distance r from the wire:

{ The electric flux at the end of the cylindrical gaussian surface is zero because the angle between E and dA is 90 (figure 3.7g).Therefore the electric flux is given by

lqin =

041k =

=

r21E

0

0

inE

qEdA == ==

0lrl)2E(dAE

rl2dA =and

or

=r

k2E where

090EdAEdAAdEE ==== orr coscos

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3.7.5 Electric field of an infinite Plane Sheet of Charge.{ Consider an infinite sheet with thin and flat surface on which there is a

uniform positive charge per unit areacharge per unit area, as shown in figure 3.7i.

z For the curved surface, the angle between E and dA is 90 thus the electric flux through the curved surface is zero.

z For the flat ends of the cylinder, both conditions are satisfied thus the electric flux through each end is

{ From the figure 3.7i, the direction of E is perpendicular to the plane and have the same magnitude at all points equidistant from the plane.

{ To find the magnitude of the electric field due to infinite plane, a small cylinder gaussian surface is drawn (figure 3.7i).

{ Both ends of gaussian surface have an area A and are equidistant from the plane.Fig. 3.7i

Er

Adr

Er

Area, A

Gaussian surface

Adr

== dAEAdEE rr AdA =andEAE =

SF027 46

therefore the total electric flux through the entire gaussian surface is

z The charge inside the gaussian surface is given by

z By applying the Gausss law,

then the magnitude of the electric field due to an infinite plane :

{ Note:z For the conductor plane sheetconductor plane sheet, the charge resides on its surface

and all the electric field lines leave on one side of the surface . Thus the total electric flux is

z By using the Gausss law, we get the magnitude of the electric field is

0E =

02E

=

EA2EAEAE =+=Aqin =

0

inE

qEdA == 0

AEA2 =

NonNon--conductor conductor plane sheetplane sheet

EAE =

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SF027 47

3.8 Electric Potential, V{ Consider a positive point charge (+q) held stationary at O exerts a

repulsive force Fe on a positive test charge (+q0) at P (figure 3.8a). A and B are two points on the line that passes through O and P.

{ The test charge at P is moved by an external force, F through a small distance dr towards A. dr is so small that the force F can be considered to be constant. Thus the work done dW by the external force is given by

Since

eFr

Fr+

q+A

0q+BPO

2r

1rr

dr

Fig. 3.8a

o0FdrdW cos= eFF =anddrFdW e=

20

e rkqqF = then dr

rkqqdW 2

0=

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{ The total work done W in bringing the test charge (+q0) from B to Ais given by

{ If r1= and r2=r then the work done in bringing the test charge from infinity to point A (W ) is

=

120 r

1r1kqqW

and

BA UUW =

energy potential electric :U

or

rkqqUW 0A ==

chargepoint point with ebetween th distance :r

where

= 21

r

r 20dr

r1kqqdW

2

0A r

kqqU =

2

1

r

r0 r

1kqqW

=

1

0B r

kqqU =

orr

qq4

1UW 00

A

==where

constant ticelectrosta :k

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{ Electric potential, V of a point in the electric fieldDefinition is defined as the work done in bringing positive test the work done in bringing positive test

charge from infinity to that point in the electric fcharge from infinity to that point in the electric field.ield.or

{ Since

0qWV =

then the equation above can be written as

done work :Wcharge test :0q

where

0

0

qr

kqq

V

=

orrq

41V

0

=rkqV =

rkqqW 0=

chargepoint :qchargepoint thepoint with ebetween th distance :rwhere

space free ofty permittivi :0) .( 212120 mNC10x858

=

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{ Electric potential is a scalar quantity.{ The S.I. unit for electric potential is the Volt (V)Volt (V) or J CJ C--11.{ The total electric potential at a point in space is equal to the algebraic algebraic

sumsum of the constituent potentials at that point.{ Note :

z The theoretical zerozero of electric potentialelectric potential of a charge is at at infinityinfinity.

z The electric potential energy of a positively charged particle increases when it moves to a point of higher potential.

z The electric potential energy of a negatively charged particle increases when it moves to a point of lower potential.

z Since charge q can be positive or negative, the electric potential can also be positive or negative.

z If the value of work donework done is negativenegative work done by the work done by the electric force (system).electric force (system).

z If the value of work donework done is positivepositive work done by the work done by the external force or on the systemexternal force or on the system.

z In the calculation of V, the sign of the charge mustsign of the charge must be substituted in the equation of V.

VqUW 0==

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{ Example 10 :Figure below shows a point A at distance 10 m from the positive point charge, q=5C.

Calculate the electric potential at point A and describe the meaning of the answer.(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)

Solution: q=5 C, r=10 mBy applying the equation of the electric potential at a point,

)())(.(

10510x09

rkqV

9

A ==19

A CJV10x54V= @ .

Meaning : Meaning : 4.5 x 104.5 x 1099 joule of work is done in bringing 1 C positive joule of work is done in bringing 1 C positive charge from infinity to the point A.charge from infinity to the point A.

++q A

m10

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{ Example 11 :Two point charges, q1=+0.3 C and q2=-0.4 C are separated by a distance of 6 m as shown in figure below.

Calculate a. the electric field strength andb. the electric potentialat point A ( 3 m from the charge q1).(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)Solution: q1=+0.3 C, q2=-0.4 C

a. By applying the equation of electric field strength, the magnitude of

E at point A.Due to q1 :

++ -- 2q1q A

m6

++ -- 2q1q A

m3r1 = m3r2 =1AE

r 2AEr

2

9

21

11A 3

3010x09rkqE

)().)(.(==

181A CN10x3E

= Direction : Direction : to the right (qto the right (q22))

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SF027 53

Due to q2 :

therefore the electric field strength at point A due to the charges is given by

b. By applying the equation of electric potential, the value of V at point A is

2

9

22

22A 3

4010x09rkqE

)().)(.(==

182A CN10x4E

= Direction : Direction : to the right (qto the right (q22))

2A1AA EEErrr +=

2A1AA VVV +=

88A 10x410x3E +=

r

18A CN10x7E

= r Direction : Direction : to the right (qto the right (q22))

+=+=

2

2

1

1

2

2

1

1A r

qrqk

rkq

rkqV

( )

+

=3

4033010x09V 9A

...

V10x3V 8A =

SF027 54

{ Example 12 :Two point charges, q1=+12 nC and q2=-12 nC are separated by a distance of 8 cm as shown in figure below.

Determine the electric potential at point P( 6 cm from the charge q2).(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)

Solution: q1=+12x10-9 C, q2=-12x10-9 C

1q ++ -- 2q

P

m8 c

m6 c

1q++ --

2q

P

m10x8 2

m10x6r 22 =m10x10r

21

=

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SF027 55

By applying the equation of electric potential, the value of V at point P is

{ Example 13 : (exercise)Four point charges are located at the corners of a square that is 8.0 cm on a side. The charges, going in rotation around the square, are q, 2q, -3q and 2q, where q = 4.8 C as shown in figure below.

2P1PP VVV +=

+=+=

2

2

1

1

2

2

1

1P r

qrqk

rkq

rkqV

V720VP =

q q2

q2 q3

cm8 Find the electric potential at the centre of the square.(Given 0=8.85 x 10-12 C2 N-1 m-2)Ans. : 1.53 x 106 V.

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3.8.1 Potential Difference{ Potential difference between two points in an electric field,

Definition is defined as the work done in bringing a positive test the work done in bringing a positive test charge from a point to another point in the electriccharge from a point to another point in the electricfield.field.

{ From the figure 3.8a, the potential difference between point A and B, VAB is given by

0

BAAB q

WV = BAAB VVV =and

0

BABA q

WVV =or

A.point toBpoint fromcharge test positive bringingin done work :BAWwhere

Apoint at potential electric :AVBpoint at potential electric :BV

charge test :0q

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SF027 57

{ Note :z If the positive test charge moving from point A to point B, thus the

potential difference between this points is given by

therefore

0

ABABBA q

WVVV ==

B.point A topoint fromcharge test positive bringingin done work :ABW

where

Apoint and Bpoint between difference potential :BAV

BAAB VV =

SF027 58

{ Example 14 :Two point charges q1=+2.40 nC and q2=-6.50 nC are 0.100 m apart. Point A is midway between them, point B is 0.080 m from q1 and 0.060 m from q2 as shown in figure below.

(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)

Solution: q1=+2.40x10-9 C, q2=-6.50x10-9 C, r1A=r2A=0.050 m, r1B=0.080 m , r2B=0.060 m

a. By applying the equation of electric potential, the value of V at point A is A2A1A VVV +=

1q++ --

2q

B

A

m060.0 m080.0

m050.0 m050.0

V738VA =

Find a. the electric potential at point A,b. the electric potential at point B,c. the work done by the electric field

on a charge of 2.5 nC that travels from point B to point A. (Young & freedman,pg.900,no.23.21)

A2

2

A1

1A r

kqrkqV +=

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SF027 59

b. By applying the equation of electric potential, the value of V at point B is

c. Given q0=2.50x10-9 CThe work done in bringing charge, q0 from point B to point A is given by

{ Example 15 :A test charge q0=+2.3x10-4 C is 5 cm from a point charge q. A work done of +4 J is required to overcome the electrostatic force to bring the test charge q0 to a distance 8 cm from charge q.Calculate :a. the potential difference between point 8 cm and 5 cm from the point

charge, q.b. the value of charge q.

B2B1B VVV +=

V705VB =B2

2

B1

1B r

kqrkqV +=

AB0BA VqW =

J10x258W 8BA= .)( BA0BA VVqW =

SF027 60

c. the magnitude of the electric field strength for charge q0 at point 5 cm from the charge q.(Given Coulombs constant, k = 9.0 x 109 N m2 C-2)

Solution: q0=+2.30x10-4 C

a. Given WAB= +4J, From the figure above, rA= 5x10-2 m, rB= 8x10-2 mBy applying the equation of potential difference, the value of VBA is

b. The electric potential at point A due to point charge, q :

)()(2

9

AA 10x5

q10x9rkqV ==

0

ABBA q

WV = V10x741V 4BA .=

q10x81V 11A .=

qBA

m10x5 2 m10x8 2

eFr

Fr

SF017

SF027 61

The electric potential at point B due to point charge, q :

The potential difference between point A and B is

c. By using the equation of electric field strength, thus

V10x741VV 4BAAB .==BAAB VVV =

C10x582q 7= .q10x1251q10x8110x741 11114 ... =

2A

A rkqE =

and

15A C N10x299E

= .

q10x125110x8

q10x9rkqV 112

9

BB .)(

)( ===

SF027 62

3.8.2 Relation Between V and E{ Consider a positive test charge, q0 placed near a positive point

charge, q. To move q0 towards q by a small displacement (r), work done (W) must be expended as shown in figure 3.8b.

{ The work done by the external force F is given byando0rFW cos=

VqW 0=

eFrF

r

rq ++ 0q+

Fig. 3.8b

rqFV

0

e =eFF =

rFW e=Since then

rVE

=

and EqF

0

e =

rEV = or

difference potential :Vwheree)nt(distancdisplacemein change :r

strength field electric :E

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SF027 63

{ In the limit when r approaches zero,

z The negative sign indicates that the value of electric potentialdecreases in the direction of electric field.

z is known as the electric potential gradient. It can be obtained from the gradient of a V against r graph.

{ An alternative unit for electric field strength, E is volts per meter where

{ The electric field produced by a pair of flat metal plates, one of which is earthed and the other is at a potential of V is uniform. This can be shown by equally spaced lines of force in figure 3.8c.

drdVE =

11 mV1CN1 =

drdV

=

rVE

0rlimit

SF027 64

z The V against r graph for pair of flat metal plates can be shown in figure 3.8d.

z From the figure 3.8d,{ The graph is a straight line with negative constant gradient,

thus

V+

Fig.3.8cFig.3.8c

d

0V =

r

V

0

V

dFig. 3.8d

)()(

0dV0

rVE

==

dVE = or EdV = Uniform Uniform EE

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SF027 65

3.9 Equipotential Surface{ Definition is defined as the locus of points that have the same the locus of points that have the same

electric potential.electric potential.{ Figures 3.9a and 3.9b are example of the equipotential surface.

z The dashed lines represent the equipotential surface (line).z The equipotential surfaces (lines) always perpendicular to the

electric field lines passing through them.

Fig. 3.9a : a uniform electric field produced by an infinite sheet of charge

Er

CA

B

Fig. 3.9b:a point charge

Er

C

AB

SF027 66

{ From the figures,then the work done to bring a test charge from B to A is given by

{ Example 16 : (exercise)At a certain distance from a point charge, the magnitude of the electric field is 500 V m-1 and the electric potential is -3.00 kV. Calculatea. the distance to the charge.b. the value of the charge. (Serway & Jewett,pg.788,no.17)(Given 0=8.85 x 10-12 C2 N-1 m-2)Ans. : 6.00 m, -2.00 C

CBA VVV =

0WBA =)( BA0AB0BA VVqVqW ==

No work is done in moving a charge No work is done in moving a charge along an along an equipotentialequipotential surface.surface.

unit 3 - electrostatics

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