22-1: what is physics? - an-najah staff · general physics ii (22102) dr. iyad saadeddin along...

General physics II (22102) Dr. Iyad SAADEDDIN

Dr. Iyad SAADEDDINChapter 22: Electric FieldsIn this chapter we will cover

The Electric Field

Electric Field Lines

Electric Field due to point charge

Electric Field due to a Continuous Charge Distribution

Motion of a Charged Particle in a Uniform Electric Field

A Dipole in an Electric field

22-1: What is physics?In previous chapter, we discussed the electric force on

charged particles when they placed near each other;Ther will be attarction or repulsion between the charges exist

in the system.The Question is: How do any charge feel the presence of

other charge without touching each other?The answer: Any charged particle 1 set up an electric field in

the space surrounding itself. If we place other charged particle 2 at any given point in that space, particle 1 “knows” of the presence of second particle 2 because particle 2 is affected by the electric field exist at that point from particle 1.

particle 1 pushes or pulls on particle 2 not by touching it but by means of the electric field produced by particle 1.

Our goal in this chapter is to define electric field and discuss how to calculate it for various arrangements of charged particles.

22-2: The Electric Field (E)Electric field E exists in a region of space surrounding a charged object.If another charged object enters a region where an electrical field is

present it will be subject to an electrical force.

Assume a small +ve test charge q0 is placed at point P in the electric field of positively charged object

The direction of at point P will be the same direction of the force excerted on the test charge q0 placed at point P

Er

Fr

Direction defined as the direction of the electrical force exerted on a small +ve charge (q0) placed at that location.

- - -- - - -- - -

E+ + ++ + + +

+ + +

Eq0

q0

22-2: The Electric Field (E)

The electric field is generally changes with position (location)The electric field is vector quantity : magnitude and direction.

Electric field points away from a +ve charge

Electric Field points towards a -ve charge

+Q -Q


0qFEr

r=

Hence, the electric field is defined by the electric force exerted on a small charged particle (q0) from an other charged object.

Er

Fr

magnitude of electric field is

0qFE =

With direction same as force direction acting on small +ve test charge q0

22-2: The Electric Field (E)

( SI unit is N/C)

A convenient way to visualize electric field patterns is to drawlines in the direction of the electric field.Such lines are called electric field lines.Remarks:

Electric field vector, , is tangent to the electric field lines at each point in space.

The number of lines per unit area through a surface perpendicular to the lines is proportional to the strength of the electric field in a given region.

Er

E is large when the field lines are close together and small when far apart.

near the sphere > far away from the sphere

Er

Er

22-3: Electric Field Lines

Remarks – continued: Field lines start from positive charges an end at negative ones.Number of lines leaving (or approaching) the charge is

proportional to the magnitude of the chargeField lines can not cross.

Examples on E-field lines:1) Electric field lines of single positive (a) and negative (b) charges.


2) two point charges

Dipole: two opposite charges of equal magnitude

two similar charges of equalmagnitude (EA>EB>EC=0)

two opposite charges of different magnitude: numberof lines leaving +2q > number of lines entering -q



3) plates

Positively charged plate: E-field lines away from plate

two oppositly charge plates: parallel lines E-field awayfrom +ve plate towards –veplate

negatively charged plate: E-field lines towards plate


To find the electric field due to a point charge q (or charged particle) at any point a distance r from charge q, we put a positive test charge q0 at that point; Calculate E from F

rrqk

qFE ˆ2

0

==r

r

Er

rr

qqkF ˆ20=

rwhere

If q is +ve E is directed away from q

If q is +ve E is directed towards q

22-4: The Electric Field due to a point charge

∑∑ =+++==i i

ii

ii r

qkEEEEE 221 ....rrrrr

For a groupe of charges, to find the E-field at point P, we can use the superposition principle (vector sum of E-fields for all charges)

with E-field magnitude2r

qkE =

Ex: A charge q1 = 7 µC is located at the origin, and a second charge q2 =- 5.0 µC is located on the x axis, 0.3 m from the origin. Find the electric field at the point P, which has coordinates (0, 0.4) m.

22-4: The Electric Field due to a point charge: Example

CNjiE

ji

jEiEE

CNjE

/)ˆ104.1ˆ101.1(

ˆ)54(108.1ˆ)

53(108.1

ˆsinˆcos

/)ˆ109.3(

552

55

222

51

×−×=

×−×=

−=

×=

r

r

r

θθ

We can calculate the vectors of E-fields

CNjijEiEEEE

isPatfieldEtotalthehence

yx /)ˆ105.2ˆ101.1(ˆˆ 5521 ×+×=+=+=

−

∑∑rrr

CNEEE yx /017.2 522 ×=+= °== − 66tan 1

x

y

EE

φ

22-4: The Electric Field due to a point charge: Example – continued from previous slide

The magnitude: The direction:


along dipole axis, at point P (at distance z from dipole center). Since we have point charges

An electric dipole: two charges of same magnitude and opposite signs (q and –q) separated by a distance d.

For the dipole shown, we can calculate E-field,

22-4: The Electric Field due to an Electric Dipole

22)(22)()2()2( )()(

dzqk

rqkEanddz

qkrqkE

+==

−==

−+

−+

The total E-field at P is the vector sumdirection-z in the

)2()2( 22)()( dzqkdz

qkEEE+

−−

=+= −+

rrr

direction-z in the )21(

1)21(

1222 ⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+−

−=⇒

zd

zdz

kqE

forming a commondenominator

22-4: The Electric Field dueto an Electric Dipole

For far distance z z>>d/2 d/2z ≈ 0

Where p = qd is the magnitude of electricdipole moment vector pr

charge ve tocharge ve- from directed always

isat vector tha is

+

pr

The E-field due to an electric dipole at point P that lie on axis pass through the dipole center and vertical to dipole axis can be calculated using same method

For z>>d/2z2 + (d/2)2 ≈ z2

22-4: The Electric Field due to an Electric Dipole

( ) )()(

)(

Because ,2/ 222)()( −+

+

=+

=== −+ rrdzqk

rqkEE

⇒−+

only axis-on x is field-E Hence,

cancels of components - z the )()( E and Err

( ) ( )( ) 2122

22

)()()(

2/2

2/2

cos2coscos

dz

d

dzqkE

EEEE

++=

⇒=+= +−+ θθθ

In the Electric dipole of previous slide, if the electric dipole has a charge magnitude of 5 µC and it is seperated be 2 cm. what is the electric force on a test charge, q1 = 2 µC, located at point P (50 cm from the dipole center)?

22-4: The Electric Field due to an Electric Dipole: Example

31

31

1

3

zqdqk

zpqkEqF

zpkE

===⇒

=⇒

Ni.F

nx-directioEN.

F

-

-

)ˆ10441(

)( ofdirection in the10441)1005(

)102)(105)(102(1099.8

2

2

32-

-2-6-69

×=

×=

××××

×=

r

z>>d/2


22-4: E - Field due to an Electric Dipole: Example -Electric dipole and atmospheric sprites

Sprites are huge flashes that occur far above a large thunderstorm ( عاصفة(برق) They produced when especially powerful lightning .(رعدية occurs between the ground and storm clouds.

This is due to E-field arising from electric dipole of earth and cloud. The E-field at high atmosphere cross critical value (Ec) and hence motion of electrons between atoms ionization light occurs in shapes of Sprites (اشباح)

When lightning (برق) occurs, Large amount of –vecharge (-q) transferred from ground to the cloud +vecharge (+q) on earth

22-4: E - Field due to an Electric Dipole: Example -Electric dipole and atmospheric sprites – continued from previous slide

by assuming a vertical electric dipole that has charge -q at cloud height h and charge +q at below-ground depth h. If q =200 C and h =6 km, what is the magnitude of the dipole’s electric field at altitude z1 =30 km from ground somewhat above the clouds) and altitude z2 =60 km (somewhat above the stratosphere) ?

333

422zqhk

zqdk

zpkE ===

* At z=30 km = 30000 m E =1.6×103 N/C at this hight, E<Ec no sprite will occur

* At z=60 km = 60000 m E =2×102 N/C at this altitude, E>Ec ionization of

atoms will occure sprites will occur

If we have a continuous charge distribution as shown

Electric Field of a continuous charge distribution

We can choose an arbitrary element charge ∆q (fraction of total charge q) the E-field (∆E - fraction of total E-field) due to ∆q at point P located at r is

The total E-field due to all charge element ∆qi at P is approximately

Electric Field of a continuous charge distributionFor charge elements approaching zero (∆qi 0 ∆qi ≡ dq ) we

will have a continuous distribution for charge qthe total field at P is

or the E-field produced at P by dq is

Note that or is a vector in the radial direction ( ) it may have components when considering other coordinate system like Cartesian (xyz) coordinate system.

dq


Electric Field of a continuous charge distributionTo perform integral calculations, it is useful to use the concept of charge density

Q = ρV dQ = ρdV

Q = σA dQ = σdA

Q = λL dQ = λdL

22-6: Electric Field due to a line charge: charged rode

Ex: A rod of length l has a charge density λ and a total charge Q. Calculate the electric field at a point P that is located along the long axis of the rod and a distance a from one end:

But

For charge segment dq we have E-field

Since has only one component in the –ve x-direction

The magnitude of is

Hence,

Ex: A ring of radius R and uniform charge density λ. Calculate the electric field due to the ring at a point P lying a distance z from its center along the central axis perpendicular to the plane of the ring (x-y plane).

22-6: Electric Field due to a line charge: charged ring

1) For sigment length (arc ds) the E-field ( ) has two components:

and

Parallel to z-axis

perpendicular to z-axis

2) If we chose other segment length ds’in front of first one as shown components from both segments cancels only z-components add

For all the ring, the perpendicular componentsof the E-field cancel

We will only have z - component for the E-field

z - component

with

and

In the z - direction

22-6: Electric Field due to a line charge: charged ring – continued from previous slide


22-6: Electric Field due to a line charge: charged ring – continued from previous slide

Integrate both sides

2πR is the circumference (length) of the ring λ(2πR) = q

In the z - direction (perpendicular to ring x-y plane)

22-6: Electric Field due to a line charge: charged arcEx: A plastic rod having a uniformly distributed charge - Q. The rod has been bent in a 120° circular arc of radius r as shown. In terms of Q and r. what is the electric field due to the rod at point P (at the origin - center of the arc) ?

We can choose other symmetric element ds’ dEycancels and dExadd as shown.

We only have x-components for the E-field

22-6: Electric Field due to a line charge: charged arc - continued

The E-Field in the x-direction for the arc is:

Note that the arc length ds = r dθ

22-7: Electric Field due to a line charge: Charged disk

Ex: A disk of radius R and uniform charge density σ. Calculate the electric field due to the disk at a point P lying a distance z from its center along the central axis perpendicular to the plane of the ring (x-y plane).

1) We can choose a charge element to be a ring of radius rarea dA = 2πrdr dq = σdA = σ (2πrdr)

2) The E-field due to the ring of charge dq is


22-7: Electric Field due to a line charge: Charged disk – continued from previous slide

Integrate for the variable r (from 0 - R)

It can be integrated with the help of integral rule:

Solve the limits and rearrange

22-8: A Point Charge In an Electric Field

E-Field

Er

Force

EqFrr

=

Acceleration

mEq

mFa

rrr

==

When a particle of charge q and mass m is placed in E-field an electric force Fe will exert on the particle causing it to accelerate

Using Newtom2nd law

If E-field is uniform a is constant equation of motion for constant a (general physics 1)

if q is +ve acceleration will have same direction of E-field

particle will move in the direction of E-field

if q is -ve acceleration will have opposite direction of E-field

particle will move in the opposite direction of E-field

Ex: A positive point charge q of mass m is released from rest, near the +ve plate, in a uniform electric field E directed as shown. Find the speed of the particle when reaching –ve plate.

Or other solution

22-8: A Point Charge In an Electric Field: Example: Charged particle in an E-field

Ex: accelerated electron: An electron enters the region of a uniform electric field as shown, with vi = 3×106 m/s and E = 200 N/C. The horizontal length of the plates is l = 0.1 m. Find a) the acceleration of the electron while it is in the electric field. b) If the electron enters the field at time t = 0, find the time at which it leaves the field c) If the electron position as it enters the field is (0,0), what is its vertical position when it leaves the field

22-8: A Point Charge In an Electric Field: Example: Charged particle in an E-field


constant== ix vv

22

210

21 t

meEtatvy yiy −=+=∆

xxx v

lv

xttvx =∆

=→=∆

No force in x-direction no acceleration in x-direction

a)

b) In the x-direction

(Accelerating downward)

c)

22-8: A Point Charge In an Electric Field: Example: continued from previous

In material, electrons and protons are connected by an electric force forming electric dipoles within the material like the water or dielectric materials.

If you consider an electric dipole with distance d between them an electric dipole moment (عزم قطبي آهربائي) exist between them

qqp +− tofromdirectedvectoraisr

22-9: Electric dipole in an electric field

pr

qdpp ==rwhere

22-9: Electric dipole in an electric fieldWhen a dipole is placed in uniform

electric field electric force F = qE will exert on both charges on different directions net force on the dipole = 0. But net force produces a net torque τ because forces acts on (عزم دوران)different lines of action about their center of mass.

θθτ

θθτ

sinsinis(dipole) chargesbothontorquetotal

sin2

sinchargeeachfor

pEqdE

qEdrF

tot ==⇒

==⇒

net torque can be expressed as

since E-force is conservative and hence it’s work store potential energy Rotation of dipole store a potential energy within dipole.

The dipole has its least potential energy (U = 0) when its moment p is lined up (┴) with the field E(when the angle θ is 90°). The potential energy U of the dipole at any other value of θ can be found by calculating the work W done by the field on the dipole when the dipole is rotated to that value of θ from 90°.

Which (work done on dipole by E-field) can be expresed as

22-9: Electric dipole in an electric field

and

UUUUUWW ffiC −=−=−=∆−== 0


22-9: Electric dipole in an electric field: Example

A neutral H2O molecule has an electric dipole moment 6.2×1030 C.m. (a) How far apart are the molecule’s centers of positive and negative charge? (b) If it is placed in an E-field of 1.5×104 N/C, what maximum torque can the field exert on it? (c) How much work must an external agent do to rotate this molecule by 180° in this field, starting from aligned position, for which θ = 0?

Solution: (a) 10 electrons and 10 protons exist in a neutral water molecule

(b) Maximum torque is when the angle between E and p

(c)

SummaryElectric Field is force per unit charge.Charges are the sources of the E-field.E-field of a charge has direction same as force direction on other +ve charge Field lines help us visualize field direction and strength.Charges under electric force will accelerateElectric dipole will have a net torque under the effect of external E-field and hence it will store potential energy

22-1: what is physics? - an-najah staff · general physics ii (22102) dr. iyad saadeddin along...

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