Chapter 21 Electric Field and Coulomb’s Law (again)

• Electric fields and forces (sec. 21.4)• Electric field calculations (sec. 21.5)• Vector addition (quick review)

C 2012 J. Becker

Learning Goals - we will learn:• How to use Coulomb’s Law (and vector

addition) to calculate the force between electric charges.• How to calculate the electric field caused by discrete electric charges.• How to calculate the electric field caused by a continuous distribution of electric charge.

Coulomb’s Law

Coulomb’s Law lets us calculate the FORCE between two ELECTRIC CHARGES.

Coulomb’s Law Coulomb’s Law lets us calculate the

force between MANY charges. We calculate the forces one at a time and

ADD them AS VECTORS. (This is called “superposition.”)

THE FORCE ON q3 CAUSED BY q1 AND q2.

Figure 21.14

SYMMETRY!

Recall GRAVITATIONAL FIELD near Earth: F = G m1 m2/r2 = m1 (G m2/r2) = m1 g where the vector g = 9.8 m/s2 in the

downward direction, and F = m g.

ELECTRIC FIELD is obtained in a similar way:

F = k q1 q2/r2 = q1 (k q2/r2) = q1 (E) where is vector E is the electric field caused by q2. The direction of the E field is determined

by the direction of the F, or the E field lines are directed

away from positive q2 and toward -q2.The F on a charge q in an E field is F = q E and |E| = (k q2/r2)

Fig. 21.15 A charged body creates an electric field.

Coulomb force of repulsion between two charged bodies at

A and B, (having charges Q and qo respectively) has magnitude:

F = k |Q qo |/r2 = qo [ k Q/r2 ] where we have factored out

the small charge qo. We can write the force in

terms of an electric field E:

Therefore we can write for

F = qo E

the electric field

E = [ k Q / r2 ]

Calculate E1, E2, and ETOTAL

at point “C”:

q = 12 nC

See Fig. 21.23: Electric field at

“C” set up by charges q1 and q1

(an “electric dipole”)

At “C” E1= 6.4 (10)3 N/C E2 = 6.4 (10)3 N/C

ET = 4.9 (10)3 N/C in the +x-direction

A

C

See Lab #2

Need TABLE of ALL vector

component VALUES.

E1

E2

ET

Fig. 21.24 Consider symmetry! Ey = 0

Xo

dq

o

dEx= dE cos =[k dq /(xo2+a2)] [xo/(xo2+ a2)1/2] Ex = k xo dq /[xo2 + a2]3/2 where xo and a

stay constant as we add all the dq’s ( dq = Q)

in the integration: Ex = k xo Q/[xo2+a2]3/2

|dE| = k dq / r2

cos = xo / r

dEx = Ex

Fig. 21.25 Electric field at P caused by a line of charge uniformly distributed along

y-axis.

Consider symmetry! Ey = 0

Xo

y

|dE| = k dq / r2dq

|dE| = k dq / r2 and r = (xo2+ y2)1/2 cos = xo/ r and cos = dEx / dE

dEx = dE cos

Ex = dEx = dE cos

Ex = [k dq /r2] [xo / r]

Ex = [k dq /(xo2+y2)] [xo /(xo2+ y2)1/2]

Linear charge density = = charge / length = Q / 2a = dq / dy

dq = dy

Ex = [k dq /(xo2+y2)] [xo /(xo2+ y2)1/2]

Ex = [k dy /(xo2+y2)] [xo /(xo2+ y2)1/2]

Ex = k xo [dy /(xo2+y2)] [1 /(xo2+ y2)1/2]

Ex = k xo [dy /(xo2+y2) 3/2]

Tabulated integral: (Integration variable “z”)

dz / (c2+z2) 3/2 = z / c2 (c2+z2) 1/2

dy / (c2+y2) 3/2 = y / c2 (c2+y2) 1/2

dy / (Xo2+y2) 3/2 = y / Xo2 (Xo2+y2) 1/2

Ex = k xo -a [dy /(xo2+y2) 3/2]

Ex = k(Q/2a) Xo [y /Xo2 (Xo2+y2) 1/2 ] -aa

Ex = k (Q /2a) Xo [(a –(-a)) / Xo2 (Xo2+a2) 1/2 ]

Ex = k (Q /2a) Xo [2a / Xo2 (Xo2+a2) 1/2 ]

Ex = k (Q / Xo) [1 / (Xo2+a2) 1/2 ]

a

Fig. 21.47 Calculate the electric field at the proton caused by the distributed

charge +Q.

Tabulated integral: dz / (c-z) 2 = 1 / (c-z)

b

is uniform (= constant)

+Q

Fig. 21.48 Calculate the electric field at -q caused by +Q, and then the force on –q:

F=qE

Tabulated integrals: dz / (z2 + a2)3/2 = z / a2 (z2 + a2) ½ for calculation of Ex

z dz / (z2 + a2)3/2 = -1 / (z2 + a2) ½

for calculation of Ey is uniform (= constant)

An ELECTRIC DIPOLE consists of a +q and –q separated by a distance d.

ELECTRIC DIPOLE MOMENT is p = q d

ELECTRIC DIPOLE in E experiences a torque:

= p x E

ELECTRIC DIPOLE in E has potential energy:

U = - p E

Fig. 21.32 Net force on an ELECTRIC

DIPOLE is zero, but torque () is into the page.

= r x F = p x E

ELECTRIC DIPOLE MOMENT is p = qd

see www.physics.sjsu.edu/Becker/physics51

Review

Vectors are quantities that have both magnitude and direction.

An example of a vector quantity is velocity. A velocity has both magnitude (speed) and direction, say 60 miles per hour in a DIRECTION due west.

(A scalar quantity is different; it has only magnitude – mass, time, temperature, etc.)

A vector may be decomposed into its x- and y-components as shown:

2 2 2

cos

sinx

y

x y

A A

A A

A A A

Note: The dot product of two vectors is a scalar quantity.

cos x x y y z zA B AB A B A B A B

The scalar (or dot) product of two vectors is defined as

sinA B AB

The vector (or cross) product of two vectors is a vector where the direction of the vector product is given by the right-hand rule.

The MAGNITUDE of the vector product is given by:

Right-hand rule for DIRECTION of vector cross product.

PROFESSIONAL FORMAT