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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.

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chapter 21 electric field and coulomb s law again
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

slide2

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

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

slide4

Coulomb’s LawCoulomb’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.

slide6

Recall GRAVITATIONAL FIELD near Earth:F = G m1 m2/r2 = m1 (G m2/r2) = m1g where the vectorg = 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)

slide7

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 = qoE

the electric fieldE = [ k Q / r2 ]

slide8

E1

See Fig. 21.23:Electric field at “C” set up by charges q1 and q1

C

ET

Calculate E1, E2, and ETOTALat point “C”:

E2

See Lab #2

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

q= 12 nC

A

Need TABLE of ALL vector component VALUES.

(an “electric dipole”)

slide9

Fig. 21.24 Consider symmetry! Ey= 0

dq

ò dEx = Ex

|dE| = k dq / r2

o

Xo

cos a = xo / r

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

Ex = k xoòdq /[xo2 + a2]3/2 where xo and a stay constant as we add all the dq’s (ò dq = Q) in the integration: Ex = k xoQ/[xo2+a2]3/2

slide11

y

Consider symmetry! Ey= 0

dq

|dE| = k dq / r2

Xo

Fig. 21.25 Electric field at Pcaused by a line of charge uniformlydistributed along y-axis.

slide12

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

dEx = dE cos a

Ex = ò dEx = ò dE cos a Ex = ò [k dq /r2] [xo / r]

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

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

dq = l dy

slide13

Ex = ò[k dq /(xo2+y2)] [xo /(xo2+ y2)1/2] Ex = ò[k l dy /(xo2+y2)] [xo /(xo2+ y2)1/2] Ex = k l xoò[dy /(xo2+y2)] [1 /(xo2+ y2)1/2]Ex = k l xoò[dy /(xo2+y2) 3/2]

Tabulated integral: (Integration variable “z”)

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

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

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

slide14

Ex = k l xo -a ò[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

slide15

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

l is uniform (= constant)

+Q

b

Fig. 21.47 Calculate the electric fieldat the protoncaused by the distributed charge +Q.

slide16

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

òzdz / (z2 + a2)3/2 = -1 / (z2 + a2) ½for calculation of Ey

l is uniform (= constant)

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

slide17

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:t = p x E

ELECTRIC DIPOLE in E has potential energy:U = - p E

slide18

ELECTRIC DIPOLE MOMENT is p = qd

t = r x F t = p x E

Fig. 21.32 Net force on an ELECTRIC DIPOLE is zero, but torque (t) is into the page.

review
Review

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

slide20

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.)

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

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

slide23

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: