Electric Fields II

Electric fields are produced by point chargesand continuous charge distributions

Definition:

F qE��������������

.E

qF

is force exerted on charge by an external field

Note: q produces a field but it is not external

Example: Electric DipoleExample: Electric Dipole

Derive an expression for at point B.E

y

a a

x

B : (0,y)

A : (x,0)

-q +q

Dipole solution:

For fun: find E at point A, and show that it is approximately proportional to x 3, at large distances x.

E = ?+++

++ + ++ + +

+

Source

+

Continuous Charge DistributionsContinuous Charge Distributions

•Cut source into small (“infinitesimal”) charges dq•Each produces

2

( )ˆ

e

dqdE k

r r

��������������

dE

r

dq

2e

dqdE k

r

��������������or

Steps:

•Draw a coordinate system on the diagram

•Choose an integration variable (e.g., x)

•Draw an infinitesimal element dx

•Write r and any other variables in terms of x

•Write dq in terms of dx •Put limits on the integral

•Do the integral or look it up in tables.

ExampleExample: Uniformly-Charged Thin Rod: Uniformly-Charged Thin Rod

(length L, total charge Q)

Ld?E

Charge/Length = “Linear Charge Density” = constant = Q/L

2rod rod

e

dqE dE k

r

����������������������������

++ ++ ++ ++ +

Solution:

In 2D problems, integrate components separatelyto obtain the electric field:

)(cos2 rdqkdEE xx {

x-component of E��������������

....... yy dEE

Example:Example: Uniformly-Charged Uniformly-Charged RingRing

Total charge Q, uniform charge/unit length,radius R

Find: E at any point (x, 0)

x

y

(x,0)

R

Solution:

Example:Example: Uniformly-Charged Uniformly-Charged SemicircleSemicircle

Charge/unit length, , is uniform Find: at originE

R

y

x

+

+ +

+

Solution:

SummarySummary

• Field of several point charges qi:

• Field of continuous charge distribution:

ii i

ie r

rq

kE ˆ 2

2

2

cos( )

sin( )

x e

y e

dqE k

rdq

E kr