Contact & Non-contactfolk in 1700’s had trouble wrapping brain around non-contact forces

Non-contactMichael Faraday (British: 1791-1867)

Field area around an object that affects other similar objects

Test charge charge that’s in an electric field

Planet

Satellite Proton

Electron

Satellite & electron experience field forces

• Gravitational field by Newton

• Electric field by Faraday

• Non-contact–Remember gravity?2 types of

Forces

r2

MGF =

ma

• What is the field?• a (or g)

ar2

MG=

• Non-contact–Remember gravity?–What’s the field for a charge?2 types of

Forces

F = m a

• What is the field?• a (or g)

r2

QkF =

qE

• What is the field?

• Ea

r2

MG= E

r2

Qk=

2 types of Forces

F =Eq F = force on test chargeq = charge of test chargeE = Electric Field (N/C)

Q creates the field

Two equations for electric fields

Er2

Qk=

Electric Field Lines

• Drawing Electric field lines–Rules

• lines tell direction a + test charge will go

• Out of (away from) positiveInto (toward) negative

• Lines always come out the object perpendicular to the surface.

• more lines = stronger field.

+ --

+ -

+ +

Properties of conductors in electrostatic equilibrium

1. Excess charges gather only on the outside of the surface, never inside.

2. The electric field is zero everywhere inside the conductor.

3. Electric field coming in or out of the conductor is perpendicular to the surface.

------

------

Σ E = 0 N/C

------

--

---- --

--

4. On an irregularly shaped surface, the charge will accumulate more where the radius of curvature is smallest.

This happens because the electrons will repel each other until they reach equilibrium.

Grounding An easy path for e’s to flow to or from

an infinite reservoir of e’s. Usually from the Earth to the electrical object.

Sample

Problems

• What is the electric field intensity of a 16 μC particle 5 cm away from the charge?

Er2

Qk=

E0.052

(16x10-6)(9x109)=

E=5.75x107 N/C

Sample

Problems

• What is the electric field intensity of a 16 μC particle 5 cm away from the charge?– If a 5 C test charge is placed at this μ

spot, how much force will be exerted on it?

Fr2

Qk=

F0.052

(16x10-6)(9x109)=

F=288 N

q

(5x10-6)

F = Eq = 5.75 x 107(5x10-6) = 288 N

or

Sample Problem

2

• What is the electric field at the origin for the three charge configuration? (electric field is a vector, don’t forget to find the angle)q1 = -1.00 µC r1 = 3.50 mq2 = +2.00 µC r2 = 5.00 mq3 = -1.50 µC r3 = 2.00 m

q1q2 q3

3.5 m5.00 m 2.00 m

Sample

Problems

Eq2

q1q2 q3

3.5 m5.00 m 2.00 m

= kQr2

Eq1 =kQ

r2 =

=(9x109)(2x10-6)

(9x109)(1x10-6)

(3.52) =(3.52)734 N/C

(52) = 719 N/C

Eq3 =kQ

r2 =(9x109)(1.5x10-6)

(22) = 3370 N/C

Same Direction+

-

-1920 N/CWhat is the force on a -3 μC test charge placed on the origin?

F=Eq=(-1920)(-3x10-6) = 5.76x10-2 N