chapter 21 electric field and coulomb’s law (again) coulomb’s law (sec. 21.3) electric fields...

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

• Coulomb’s Law (sec. 21.3)• Electric fields & forces (sec. 21.4 & -6)• Vector addition

C 2009 J. F. Becker

Vectors (Review) Used extensively throughout the course

INTRODUCTION: see Ch. 1 (Volume 1)

C 2009 J. Becker

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 composed of 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

C 2009 J. F. Becker

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:

C 2009 J. F. Becker

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

Right-hand rule for

DIRECTION of vector cross

product.

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.

21-9 Coulomb’s Law – vector problem

Net force on charge Q is the vector sum of the forces by the other two charges.

Coulomb’s Law -forces between two charges

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 the vector E is the electric field caused by q2. The direction of the E field is determined by the direction of the F, or as you noticed in lab #1, the E field lines are directed away from a positive q2 and toward a -q2.The F on a charge q in an E field is

F = q E and |E| = (k q2/r2)C 2009 J. F. Becker

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 points “A” & “C”:

q = 12 nC

Electric field at“A” and “C” set up by charges q1 and

q1

(an electric dipole)

a) E1= 3.0 (10)4 N/C E2 = 6.8 (10)4 N/C

EA = 9.8 (10)4 N/C

c) E1= 6.4 (10)3 N/C E2 = 6.4 (10)3 N/C

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

A

C

Lab #1

Electric field at P caused by a line of charge along the y-axis.

Consider symmetry! Ey = 0

Xo

y

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 is

constant as we add all the dq’s (=Q) in the integration: Ex = k xo Q/[xo2+a2]3/2

|dE| = k dq / r2

cos =xo / r

Calculate the electric field at +q caused by the distributed charge +Q.

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

d

b

Electric field at P caused by a line of charge along the y-axis.

Consider symmetry! Ey = 0

Xo

y

Electric field at P caused by a line of charge along the y-axis.

Consider symmetry! Ey = 0

Xo

y

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

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

Our integral=k (Q/2a) Xo 2[y /Xo2 (Xo2+y2) 1/2 ]0

a

Ex = k (Q /2a) Xo 2 [(a –0) / Xo2 (Xo2+a2) 1/2 ]

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

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

Notation change

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Calculate the electric field at -q

caused by +Q, and then the force on -q.

Tabulated integral: dz / (z2 + a2)3/2 = z / a2 (z2 + a2) 1/2 z dz / (z2 + a2)3/2 = -1 / (z2 + a2) 1/2

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 EC 2009 J. F. Becker

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.edu/becker/physics51

Review

C 2009 J. F. Becker