ELECTROSTATICS:The study of the behavior of stationary charges

ELECTRIC CHARGE

There are two types of electric charge, arbitrarily called positive and negative. Rubbing certain electrically neutral objects together (e.g., a glass rod and a silk cloth) tends to cause the electric charges to separate. In the case of the glass and silk, the glass rod loses negative charge and becomes positively charged while the silk cloth gains negative charge and therefore becomes negatively charged. After separation, the negative charges and positive charges are found to attract one another.

When a rubber rod is rubbed against fur, electrons When a rubber rod is rubbed against fur, electrons are removed from the fur and deposited on the rod.are removed from the fur and deposited on the rod.

The rod is said to beThe rod is said to be negatively chargednegatively charged because of because of anan excessexcess of electrons. The fur is said to be of electrons. The fur is said to be positively chargedpositively charged because of abecause of a deficiencydeficiency of of electrons.electrons.

Electrons Electrons move from move from fur to the fur to the rubber rod.rubber rod.

positive

negative

+ + + +

--

--

When a glass rod is rubbed against silk, electrons are When a glass rod is rubbed against silk, electrons are removed from the glass and deposited on the silk.removed from the glass and deposited on the silk.

The glass is said to beThe glass is said to be positivelypositively chargedcharged because because of aof a deficiencydeficiency of electrons. The silk is said to be of electrons. The silk is said to be negatively chargednegatively charged because of anbecause of an excessexcess of of electrons.electrons.

Electrons Electrons move from move from glass to the glass to the silk cloth.silk cloth.

positive

negative- - - -

+ +

+ +

silk

glass

Pith-ball Electroscope

Gold-leaf Electroscope

Laboratory devices used to study the existence of two kinds of electric charge.

Laboratory devices used to study the existence of two kinds of electric charge.

1. Charge the rubber rod by rubbing against fur.1. Charge the rubber rod by rubbing against fur.

2. Transfer electrons from rod to each pith ball.2. Transfer electrons from rod to each pith ball.

The two negative charges repel each other.

The two negative charges repel each other.

1. Charge the glass rod by rubbing against silk.1. Charge the glass rod by rubbing against silk.

2. Touch balls with rod. Free electrons on the balls 2. Touch balls with rod. Free electrons on the balls move to fill vacancies on the cloth, leaving each of move to fill vacancies on the cloth, leaving each of the balls with a deficiency. (Positively charged.)the balls with a deficiency. (Positively charged.)

The two positive charges The two positive charges repelrepel each each other.other.

The two positive charges The two positive charges repelrepel each each other.other.

fur

Rubber

Attraction

Note that the negatively chargedNote that the negatively charged (green)(green) ball is ball is attractedattracted to the positively chargedto the positively charged (red)(red) ball.ball.

Opposite Charges Attract!Opposite Charges Attract!

silk

glass

Like charges repel; unlike charges attract.Like charges repel; unlike charges attract.

NegNeg PosNegPosPos

Charging by Contact  

Some electrons leave rod and spread over sphere.

Charging by Induction    

Rod does not touch sphere.  It pushes electrons out of the back side of the sphere and down the wire to ground.  The ground wire is disconnected to prevent the return of the electrons from ground, then the rod is removed.

The law of conservation of electric charge: "The net amount of electric charge produced in any process is zero." Another way of saying this is that in any process electric charge cannot be created or destroyed, however, it can be transferred from one object to another.

Charged comb attracts neutral bits of paper.

Charged comb attracts neutralwater molecules.

Applications of Electrostatic Charging                                        

                               

Fine mist of negatively charged gold particles adhere to positively charged protein on fingerprint.

                                                     

Negatively charged paint adheres to positively charged metal.

TheThe quantity of chargequantity of charge ((qq)) can be defined in terms can be defined in terms of the number of electrons, but theof the number of electrons, but the Coulomb (C)Coulomb (C) is is a better unit for later work. a better unit for later work.

The Coulomb: 1 C = 6.25 x 1018 electronsThe Coulomb: 1 C = 6.25 x 1018 electrons

The charge on a single electron is:The charge on a single electron is:

1 electron: e- = -1.6 x 10-19 C1 electron: e- = -1.6 x 10-19 C

The The coulombcoulomb (selected for use with electric (selected for use with electric currents) is actually a currents) is actually a very large unitvery large unit for static for static electricity. It is common to use the metric prefixes.electricity. It is common to use the metric prefixes.

1 C = 1 x 10-6 C1 C = 1 x 10-6 C 1 nC = 1 x 10-9 C1 nC = 1 x 10-9 C

1 pC = 1 x 10-12 C1 pC = 1 x 10-12 C

COULOMB’S LAW Coulomb’s Law states that two point charges exert a force (F) on one another that is directly proportional to the product of the magnitudes of the charges (q) and inversely proportional to the square of the distance (r) between their centers. The equation is:

F kq q

r 1 2

2F = electrostatic force (N)q = charge (C)k = 9x109 N•m2/C2

r = separation between charges (m)

The value of k can also be expressed in terms of the permittivity of free space (εo):

ko

1

49x109 N. m2/C2

The proportionality constant (k) can only be used if the medium that separates the charges is a vacuum. If the region between the point charges is not a vacuum then the value of the proportionality constant to be used is determined by dividing k by the dielectric constant (K). For a vacuum K = 1, for distilled water K = 80, and for wax paper K = 2.25

Problem-Solving Strategies

1. 1. Draw Draw and and labellabel a figure indicating a figure indicating positive and negative charges along positive and negative charges along with the given distances.with the given distances.

2. Draw the force of 2. Draw the force of attractionattraction or or repulsionrepulsion on the given charge on a on the given charge on a neat, labeled FBD.neat, labeled FBD.

3. Find the 3. Find the resultant force.resultant force.

Important: Important: Do not use the signsDo not use the signs of the of the charges when applying Coulomb's charges when applying Coulomb's law!law!

11.1 Two charges q1 = - 8 μC and q2= +12 μC are placed 120 mm

apart in the air. What is the resultant force on a third charge q3 = - 4 μC placed midway between the other charges?

q1 = - 8x10-6 C

q2= +12x10-6 C

q3 = - 4x10-6 C

r = 0. 120 m

-q1

+q2

-q3

F1

F2

FR

1 31 2

q qF k

r

2 32 2

q qF k

r

0.06 m 0.06 m

= 80 N

= 120 N

FR = 80 + 120 = 200 N, to the right

9 6 6

2

9 10 (8 10 )(4 10 )

(0.06)

x x x

9 6 6

2

9 10 (12 10 )(4 10 )

(0.06)

x x x

11.2 Three charges q1 = +4 nC, q2 = -6 nC and q3 = -8 nC are

arranged as shown. Find the resultant force on q3 due to the other

two charges. q1 = +4x10-9 C

q2= -6x10-9 C

q3 = -8x10-9 C

 

F2

F1

37˚ θ

= 2.88x10-5 N

= 6.75x10-5 N

FR

9 9 91 3

1 2 2

9 10 (4 10 )(8 10 )

(0.1)

q q x x xF k

r

9 9 92 3

2 2 2 2

9 10 (6 10 )(8 10 )

(8 10 )

q q x x xF k

r x

F2

F1

37˚ θFrom the FBD:

Σ Fx = F2 - F1 cos 37˚

= (6.75x10-5) - (2.88x10-5)(cos 37˚)= 4.45x10-5 N

Σ Fy = F1 sin 37˚

= (2.88x10-5)(sin 37˚)= 1.73x10-5 N

F x x ( . ) ( . )4 45 10 173 105 2 5 2 = 4.8x10-5 N

tan.

.1

5

5

173 10

4 45 10

x

x= 21˚

FR (4.8x10-5 N, 21˚)

FR

ELECTRIC FIELD

An electric field is said to exit in a region of space in which an electric charge will experience an electric force. The magnitude of the electric field intensity is given by:

EF

q Units: N/C

The direction of the electric field intensity at a point in space is the same as the direction in which a positive charge would move if it were placed at that point. The electric field lines or lines of force indicate the direction. The electric field is strongest in regions where the lines are close together and weak when the lines are further apart.

11.3 The electric field intensity between two plates is constant and directed downward. The magnitude of the electric field intensity is 6x104 N/C. What are the magnitude and direction of the electric force exerted on an electron projected horizontally between the two plates?

E = 6x104 N/Cqe = 1.6x10-19 C

F = qE = 1.6x10-19 (6x104) = 9.6x10-15 N, upward

11.4 Show that the gravitational force on the electron of example 11-3 may be neglected.

me = 9.11x10-31 kg

FG = mg

= 9.11x10-31 (9.8) = 8.92x10-30 N

The electric force is larger than the gravitational force by a factor of 1.08x1015!

The electric field intensity E at a distance r from a single charge q can be found as follows:

Ekq

r

2Units: N/C

11.5 What is the electric field intensity at a distance of 2 m from a charge of -12 μC?

r = 2 mq = -12 μC

Ekq

r

2

9 10 12 10

2

9 6

2

x x( )= 27x103 N/C, towards q

When more than one charge contributes to the field, the resultant field is the vector sum of the contributions from each charge.

Ekq

r

2

Units: N/C

11.6 Two point charges q1 = -6 nC and q2 = +6 nC, are 12 cm apart,

as shown in the figure. Determine the electric field a. At point A

q1 = -6 x10-9 C

q2 = +6 x10-9 C

E1

E2

ER

Ekq

r

2

Ex x

x1

9 9

2 2

9 10 6 10

4 10

( )

( )= 3.38x104 N/C, left

Ex x

x2

9 9

2 2

9 10 6 10

8 10

( )

( )= 8.44x103 N/C, left

E1

E2

ER

ER = E1 + E2

= 3.38x104 + 8.44x103

= 4.22x104 N/C, left

11.6 Two point charges q1 = -6 nC and q2 = +6 nC, are 12 cm apart,

as shown in the figure. Determine the electric field b. At point B

q1 = -6x10-9 C

q2 = +6x10-9 C

E1

37º

E2

ER

θ

Ex x

x1

9 9

2 2

9 10 6 10

9 10

( )

( )= 6.67x103 N/C

Ex x

x2

9 9

2 2

9 10 6 10

15 10

( )

( )= 2.4x103 N/C

E1

37º

E2

ER

θ

From vector diagram:

Σ Ex = - E2cos 37˚

= - (2.4x103)(cos 37˚)= -1916.7 N/C

Σ Ey = E2 sin 37˚- E1

= (2.4x103)(sin 37˚) - (6.67x103) = - 5225.6 N/C

ER ( . ) ( . )1916 7 522562 2 = 5566 N/C

E1

37º

E2

ER

θ

tan.

.1 52256

1916 7= 70˚

180˚ + 70˚ = 250˚

ER (5566 N/C, 250˚)

GRAVITATIONAL POTENTIAL ENERGY

Consider that a mass m is moved from level A to level B. An external force F equal to the weight mg must be applied to move the mass against gravity. The work done by this force is W = mgh

When the mass m reaches level B, it has a potential for doing work relative to level A. The system has a potential energy (U) that is equal to the work done against gravity. The potential energy coverts into kinetic energy as the mass falls from level B.

ELECTRIC POTENTIAL ENERGY

When a charge q is moved against a constant electric force for a distance d, the potential energy of the system is equal to the work done to move the charge:

PE = W = F d = qE d

If the charge is released, it will acquire a kinetic energy as it returns the same distance.

 An important difference between gravitational potential energy and electric potential energy is that in the case of gravity, there is only one kind of mass, and the forces involved are always forces of attraction.

Therefore, a mass at higher elevations always has the greater potential energy relative to the earth.

This is not true in the electrical case because of the existence of the negative charge. In the figure below, a positive charge has a greater potential energy at point B than at point A. Work has been done against the electric field.

On the other hand, if a negative charge were moved from point A to B, work will be done by the field. A negative charge would have a lower potential energy at B, exactly the opposite situation of a positive charge.

Whenever a positive charge is moved against an electric field, the potential energy increases.Whenever a negative charge is moved against an electric field, the potential energy decreases.

+

-

When a positive charge is placed in an electric field:

- It moves in the direction of the field

- It moves from a point of higher potential to a point of lower potential

- Its electrical potential energy decreases and its kinetic energy increases

When a negative charge isplaced in an electric field- It moves opposite to the direction of the field- It moves from a point of lower potential to a point of higher potential- Its electrical potential energy increases and its kinetic energy increases- Work has to be done on the charge for it to move from point A to point B

+

-

ELECTRIC POTENTIAL ENERGY

The electric potential energy of a system composed of a charge q and another charge Q separated by a distance r is equal to the work done against the electric forces in moving a charge +q from infinity to that point.

PE kqQ

r Joules (J)

11.7 A charge of + 2 nC is 20 cm away from another charge of + 4 μC. a. What is the potential energy of the system?

q = +2x10-9CQ = + 4x10-6 Cr = 0.20 m

PEkQq

r

9 10 4 10 2 10

0 2

9 6 9x x x( )( )

.

= 3.6x10-4 J

b. What is the change in potential energy if the 2 nC charge is moved to a distance of 8 cm from the 4 μC charge?

r = 0.08 m

PEx x x

9 10 4 10 2 10

0 08

9 6 9( )( )

.= 9.0x10-4 J

ΔPE = 9.0x10-4 - 3.6x10-4 = 5.4x10-4 J

• If the charges have the same sign, PE is positive– Positive work must be done to force the two

charges near one another– The like charges would repel

• If the charges have opposite signs, PE is negative– The force would be attractive– Work must be done to hold back the unlike charges

from accelerating as they are brought close together

ELECTRIC POTENTIAL

The advantage of the concept of an electric field as force per unit charge was to assign an electrical property to space. If the electric field intensity is known at some point, the force on a charge placed at that point can be predicted.

It is equally convenient to assign another property to the space surrounding a charge. This will allow us to predict the potential energy due to another charged placed at any point. This property of space is called electric potential.

The electric potential V is defined in terms of the work to be done on a charge to move it against an electric field.

The electric potential V is a scalar quantity defined as the potential energy per unit charge.

Units: J/C = Volt (V)A

PEV

q

The potential at a point is defined in terms of a positive charge. The potential due to a positive charge is positive, and the potential due to a negative charge is negative.

Substitution of the equation for potential energy yields:

/A

PE kQq rV

q q

A

kQV

r Units: Volt (V)

ELECTRIC POTENTIAL due to POINT CHARGESThe potential V at a point a distance r from a charge Q is equal to the work per unit charge done against electric forces in bringing a positive charge +q from infinity to that point. The point of zero electric potential is taken to be at an infinite distance from the chargeIn other words, the potential at some point A as shown in the figure, is equal to the potential energy per unit charge.

A A

kQV

r

The potential is the same at equal distances from a spherical charge. The dashed lines in the figure below are called equipotential lines. Note that the lines of equal potential are always perpendicular to the electric field lines.

Equipotential Lines

An equipotential surface is a surface on which all points are at the same potential

No work is required to move a charge at a constant speed on an equipotential surface

These plots show the potential due to

(a) positive and

b) negative charge.

Electric Potential Due to Point Charges

11.8 a. Calculate the potential at a point A that is 30 cm distant from a charge of- 2 μC.

Q = - 2x10-6 Cr = 0.3 m V

kQ

rA 9 10 2 10

0 3

9 6x x( )

.= - 6x104 V

b. Find the potential energy if a + 4 nC is placed at point A.

q = +4x10-9 C PE = qVA

= 4x10-9 (-6x104) = -24x10-5 J

The potential in the vicinity of a number of charges is equal to the algebraic sum of the potentials due to each charge.

11.9 Two charges + 6 μC and - 6 μC are separated by 12 cm, as shown. Calculate the potential a) at point A

Q1 = + 6x10-6 C

Q2 = - 6x10-6 C

r1 = 4x10-2 m

r2 = 8x10-2 m

VkQ

r

Vx x

x

x x

xA

9 10 6 10

4 10

9 10 6 10

8 10

9 6

2

9 6

2

( ) ( )= 6.75x105 V

positive = the field does the work

b) at point B.

VkQ

r

Vx x

x

x x

xB

9 10 6 10

16 10

9 10 6 10

4 10

9 6

2

9 6

2

( ) ( )= - 1.01x106 V

negative = an external source of energy must perform the work

11.10 What is the electric potential at the center of a square of sides equal 1 m if the charges placed at the corners are +1 nC, -2 nC, +3 nC and -4 nC (read in a CW direction)

+1 -2

- 4 +3

The center of the square is equidistant from all four charges, a distance r of

31 2 4kqkq kq kq

r r r r V = V1 + V2 + V3 + V4

= - 25 V

2

d= 0.707

1 2 3 4( )k q q q qV

r

9 9

0.707

9 10 (1 2 3 4) 10x

The potential difference between two points A and B to be equal to:

In other words the potential difference between two points is the work per unit positive charge done by electric forces in moving a small test charge from the point of higher potential to the point of lower potential.

A B

WV V V

q

11.11 Consider the two charges on problem 11.8a. What is the potential difference between points A and B?

VA = 6.75x105 V

VB = -10.1x105 V

ΔV = VA-VB

= 6.75x105 - (-1.01x106) = 1.69x106 V

b. How much work is done by the electric field in moving a -2 nC charge from A to B?

q = -2x10-9 CW = q ΔV = (-2x10-9)(1.69x106) = - 3.37x10-3 J

POTENTIAL AND ELECTRIC FIELD

The potential difference between two oppositely charged plates is equal to the product of the field intensity and the plate separation.

W = q ΔV = F d = q E d

ΔV = E d

Units: ΔV = Volts (V)E = V/m or N/C

11.12 The potential difference between two plates 5 mm apart is 10 kV. Determine the electric field intensity between the two plates.

d = 5x10-3 mV = 10x103 V

EV

d

10 10

5 10

3

3

x

x= 2x106 V/m

CAPACITORS

A capacitor is a device that can store electric charge and consists of two conducting objects placed near one another but not touching. A typical capacitor consists of a pair of parallel plates of area A separated by a distance d. Often the two plates are rolled into the form of a cylinder with paper or other insulator.

If a potential difference or voltage is applied to a capacitor it becomes charged. The amount of charge acquired by each plate is proportional to the voltage.

CQ

V

C is the capacitance in farads (F), Q is the charge in (C) V is the voltage (V)

The capacitance of a parallel-plate capacitor depends on the geometry of the capacitor according to this equation:

CA

do

εo = 8.85x10-12 C2/Nm2

Where C is the capacitance (F), A is the area of the plates (m2) and d is the separation between the plates (m)

Applications of Capacitors:                 Keyboards and Defibrillators                                              

• Computers use capacitors in the keyboard:• The keyboards use capacitors at the bases of the keys• When the key is pressed, the capacitor spacing decreases

and the capacitance increases• The key is recognized by the change in capacitance

11.13 a. Find the capacitance of a capacitor whose plates are 12 cm x 12 cm separated 1 mm by an air gap.

A = 0.12 x 0.12 m2

d = 0.001 mV = 12 V

CA

do 8 85 10

012 012

0 00112.

( . )( . )

( . )x = 1.27x10-10 F

CQ

V

Q = CV = 1.27x10-10 (12) = 1.53x10-9 C

b. Find the charge on each plate if the capacitor is connected to a 12 V battery.

ENERGY OF A CHARGED CAPACITOR Consider a capacitor initially uncharged. When a source of voltage is connected to it, the voltage between the plates increases as charge is transferred.  The energy of a charged capacitor can be found with these equations:

PE QV CVQ

C

1

2

1

2 22

2

Units: Joules (J)

CAPACITORS IN SERIES AND PARALLEL

These are the symbols used in different arrangements of capacitors:

CAPACITORS IN SERIES

The concept of equivalent capacitance refers to the capacitance of a single capacitor that could replace all the capacitors in a circuit.

31 2

1 2 3

QQ Q Q

C C C C

Series capacitors always have the same charge.

The voltage across the equivalent capacitor Ceq is

the sum of the voltage across both capacitors.

CAPACITORS IN SERIES

Charge is same: series connection of capacitors.

Q = Q1 = Q2 =Q3

Battery

C1 C2C3

++

--

++

++

--

--

Q1 Q2 Q3

Total voltage (V)

series connection = Sum of voltages

V = V1 + V2 + V3

Battery

C1 C2C3

++

--

++

++

--

--

V1 V2 V3

• •A B

Equivalent Capacitance in Series: Where does the equation come from?

V = V1 + V2 + V3

Q1= Q2 = Q3

++

--

++

++

--

--

C1 C2 C3

V1 V2 V3

31 2

1 2 3

QQ Q Q

C C C C

1 2 3

1 1 1 1

eC C C C

Equivalent Equivalent CCe e for for capacitors capacitors in series:in series:

1

1 1n

ie iC C

1

1 1n

ie iC C

; Q Q

C VV C

CAPACITORS IN PARALLEL

CAPACITORS IN PARALLEL

Parallel Capacitors always have the same voltage drop across each of them.

The charge on the equivalent capacitor CEQ is the sum of the

charges on both capacitors.

Equivalent Capacitance in Parallel: Where does the equation come from?

Q = Q1 + Q2 + Q3

; Q

C Q CVV

Equivalent Equivalent CCe e for for capacitors capacitors in parallel:in parallel:

1

n

e ii

C C

1

n

e ii

C C

Equal Voltages: Equal Voltages: CVCV = C= C11VV11 + C + C22VV22 + +

CC33VV33

Parallel capacitors in

Parallel:C2

C3

C1

++

--

++

--

++

--

CCee = C = C11 + C + C22 + + CC33

11.14 a. Find the equivalent capacitance of the circuit.

C2 and C4 are in series

1 1 1

2 4 2 4C C C,

CC C

C C2 42 4

2 4,

2 4

2 4

( )= 1.33 μF

C2 = 2 μF, C3 = 3 μF, C4 = 4 μF

V = 120 V

C3 is now in parallel with C2,4

Ceq= C3 + C2,4

= 3 +1.33 = 4.33 μF

b. Determine the charge on each capacitor.

First, we need to find the total charge of the system

QT = CeqV

= 4.33 (120) = 520 μC

Q3 = C3V

= 3(120) = 360 μC

Q2 and Q4 have the same charge since they are

in series:Q2 = Q4 = QT - Q3

= 520 - 360 = 160 μC

Q3 = 360 μC, Q2 = Q4 =160 μC

c. What is the voltage across the 4 μF capacitor?

VQ

C44

4

160

4= 40 V

The remaining voltage (120 - 40 = 80 V) goes through the C2 capacitor.