medan listrik (electric fields)

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Medan Listrik (Electric fields). Hukum Coulomb  bagaimana muatan listrik berinteraksi dengan muatan listrik yang lain . Tetapi bagaimana muatan listrik tersebut ‘tahu’ ada kehadiran muatan listrik lain dan juga bagaimana mampu mendeteksi muatan tersebut negatif atau positip? - PowerPoint PPT Presentation

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Medan Listrik (Electric fields)

• Hukum Coulomb bagaimana muatan listrik berinteraksi dengan muatan listrik yang lain.

Tetapi bagaimana muatan listrik tersebut ‘tahu’ ada kehadiran muatan listrik lain dan juga bagaimana mampu mendeteksi muatan tersebut negatif atau positip?

• Maka muncul konsep medan listrik

Secara ilustratif

• Satuan medan listrik (N/C)

• Arah gaya listrik mendefinisikan juga arah medan listrik

Garis-garis gaya Listrik (Electric field lines):

• Michael Faraday (abad 19) : memperkenalkan ide medan listrik dan dipikirkan sebagai ruang di sekitar benda bermuatan listrik dipenuhi oleh garis-garis gaya

• Garis-garis gaya sebagai cara memvisualisasikan adanya medan listrik

• Arah garis-garis gaya menunjukkan arah gaya listrik yang bekerja pada tes muatan positip

• Kerapatan garis gaya sebanding dengan besarnya medan listrik

Kekuatan medan terkait dengan jumlah garis-garis yang melintasi satuan luas yang tegak lurus medan

Karena gaya listrik adalah

Kekuatan (besarnya) medan listrik:

Two positive point charges, q1=+16 μC and q2=+4.0 μC, are separated in a vacuum by a distance of 3.0 m, as Figure illustrates. Find the spot on the line between the charges where the net electric field is zero.

Examples:

SUPERPOSISI MEDAN LISTRIK

Example 5. Vector properties of Electric Fields

Two point charges are lying on the y axis in Figure a: q1=–4.00 μ C and q2=+4.00 μ C. They are equidistant from the point P, which lies on the x axis. (a) What is the net electric field at P?

MEDAN LISTRIK dari distribusi muatan listrik

Electric dipole電偶極

• In nature, many molecules carry no net charge, but there are still finite electric fields around the molecules.

• it is because the charge are not uniformly distributed, the simplest form of charge distribution is a electric dipole, as shown in the figure.

• one can calculate the electric field along the z-axis as follows:

Rewrite the electric field in the following form:

Because we are usually interested in the case where z>>d, therefore we can use an approximation to simplify E:

Throw away higher order terms of d/z,

Or with p=qd (electric dipole moment),

Note

• the dipole electric field reduces as 1/r3, instead of 1/r of a single charge.

• although we only calculate the fields along z-axis, it turns out that this also applies to all direction.

• p is the basic property of an electric dipole, but not q or d. Only the product qd is important.

E

E

p

Electric field due to a line charge

• we now consider charges uniformly distributed on a ring, rather than just a few charges.

• again we use the superposition principle of electric fields, just as what we have done on electric dipoles.

• assume the ring has a linear charge density λ.

• then for a small line element ds, the charge is

So it produces a field at point P of:

Or rewrite as

• Here we only consider the field along z-axis, dEcosθ, because any fields perpendicular to z direction will be cancelled out at the end.

• since

Sum over all fields produced by other elements on the ring:

An infinite long line charge

204 r

dldE

tan and cos

RlR

r

dRdl 2sec

R

ddE

04

120

00

sinsin4

sin4

cos4

2

1

2

1

R

Rd

RE

For an infinite line,

2/21

RE

04

2

A point charge in an electric field

By the definition of electric field, a point charge will experience a force equal to:

The motion of a point charge can now be described by Newton’s law.

Robert A. Millikan, in 1909, made use of this equation to discover that charge is quantized and he was even able to find the value of fundamental charge e=1.6x10-19C.

The electric field between the plates are adjusted so that the oil drop doesn’t move:

Emgq

mgqE

/

He found that q=ne, always a multiple of a fundamental charge e.

mm66.0

2

1

2

12

2

xv

L

m

QEaty

m

QEa

maQEF

Dipole in an electric field

• The response of an electric dipole in an electric field is different from a charge in an electric field.

• Since a dipole has no net charge, so the net force acting on it must be zero.

• however, each charge in the dipole does experience forces from the field.

• the forces are equal in magnitude but opposite in direction, and so there is net torque

sinFd dF

sin

sin

pE

qEd

Or in vector form

Ep

p is the dipole moment 偶極矩

Therefore, a dipole in an uniform electric field experiences a torque that is proportional to the dipole moment, and does not depends on any detail of the dipole.

Ep

Potential energy

Under the effect of the field, the dipole will now oscillates back and forth, just like a pendulum under the effect of gravitation field.

The motion of the dipole of cause requires energy. The work done on dipole by the torque is dW

Similar to gravitation potential, we can define a electric potential energy of the dipole in an electric field :

In the vector form, the potential energy is simply a dot product of the dipole moment and the field.

The potential energy is the lowest when the angle = 0 (equilibrium position) and has the largest values when the angle = 180°.

If there is energy loss to the surroundings, the oscillation will die out gradually unless energy is pumped in continuously from the field (an oscillating field).

This is essentially the principle of a microwave oven.

• Water molecule has large dipole moment of 6.2x10-30Cm.

• the dipoles vibrate in response to the field and generate thermal energy in the surrounding medium

• materials such as paper and glass, which has no dipoles, do not become warm

• The large dipole moment of water molecule attracts Na and Cl ions and breaks the ionic bond between these ions.

• Therefore, salt dissolves easily in water.

Note: A dipole in an uniform field experiences no net force, but it does in a non-uniform field.

F+>F-

Induced dipole

Charges in the comb produce a non-uniform field

)2/(4

22

0 a

eE

Ep (a) τ=0

(b) τ=pE

(c) τ=0

0

)'(4

'

)(4 2/3220

2/3220

RD

DQ

RD

QDE

the field produced by the rings is

Q

QRD

RDQ

2/3

2/3

22

22

5

13

''

15363 1087.8)8.9()106.0(3

4)1000(

3

4 grmg(a)

(b)

e

q

mgqE

120

109.1462/1087.8 1715

)90cos()90cos( if

ifif

pE

EpEpUU

電四極

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