Electric Electric FieldsFields

What is an electric What is an electric field???field???

It is an area of influence around a It is an area of influence around a charged object.charged object.

Positively charged objects within Positively charged objects within that field will experience an that field will experience an electrical force.electrical force.

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Electric Field

Test Charge Electric

Field

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Calculating the magnitude of Calculating the magnitude of an electric fieldan electric field

The magnitude is proportional to the The magnitude is proportional to the amount of electrical force exerted on amount of electrical force exerted on the positively charged object.the positively charged object.

Electric Field Strength = electric force/test Electric Field Strength = electric force/test charge charge

E =E = FFelectricelectric/q/q00

SI Unit = N/CSI Unit = N/C

Sample ProblemMr. Luna pulls a wool sweater over his head, which charges his body as the sweater rubs against his cotton shirt. What is the electric field strength at a location where a 1.60 x 10-19 C piece of lint experiences a force of 3.2 x 10-9 N as it floats near Mr. Luna?qq00 = 1.60 x 10 = 1.60 x 10-19-19

CCF = 3.2 x 10F = 3.2 x 10-9-9 NN

E = E = FFelectricelectric/q/q00

= 3.2 x 10= 3.2 x 10-9-9 N / 1.60 x N / 1.60 x 1010-19-19 C CE = 2.0 x 10E = 2.0 x 101010 N/CN/C

Calculating the strength of Calculating the strength of an electric fieldan electric field

The strength of an electric field depends The strength of an electric field depends on: on:

1. the charge produced by the electric field1. the charge produced by the electric field

2. the distance between the test charge 2. the distance between the test charge and the and the

electric fieldelectric field

Electric Field Strength = Coulomb constant x field charge/ Electric Field Strength = Coulomb constant x field charge/ (distance)²(distance)²

E = (Kc x q)/r²E = (Kc x q)/r²Kc = 8.99 x 10 Kc = 8.99 x 10 9 9 N • m² / C² N • m² / C²

Sample ProblemA fly accumulates 3.0 x 10-10 C of positive charge as it flies through the air. What is the magnitude and direction of the electric field at a location of 2.0 cm away from the fly?

E = (Kc x q)/r²E = (Kc x q)/r²

q = 3.0 x 10q = 3.0 x 10-10-10 C C

Kc = 8.99 x 10Kc = 8.99 x 109 9 N • m² / C² N • m² / C²

r = 0.02 r = 0.02 mm

== (8.99 x 10(8.99 x 109 9 N•m²/C²)N•m²/C²) (3.0 x (3.0 x 1010-10-10 C) C)

(0.02 m)²(0.02 m)²E = 6743 N/C away from E = 6743 N/C away from the flythe fly

Electric Field LinesElectric Field Lines Represent the magnitude and Represent the magnitude and

direction of an electric field with direction of an electric field with respect to the positive test charge.respect to the positive test charge.

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More arrows = greater magnitude

When two point charges of equal magnitude, When two point charges of equal magnitude, but opposite signs interact with each other, but opposite signs interact with each other, they form an they form an ““electric dipole”electric dipole”..

Electric Field LinesElectric Field Lines

Like charged objects that interact Like charged objects that interact with each another, have their field with each another, have their field lines moving away from one another.lines moving away from one another.

Electric Field LinesElectric Field Lines

When interacting charges are of a When interacting charges are of a different magnitude, only a fraction different magnitude, only a fraction of the field lines from the greater of the field lines from the greater charge will move to the lesser charge will move to the lesser charge. charge.

Electric Field LinesElectric Field Lines