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September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 1
Electric Field from an Electric Dipole
A system of two oppositely charged point particles is called an electric dipole e vector sum of the electric eld from the two charges gives the electric eld of the dipole We have shown the electric eld lines from a dipole We will derive a general expression good anywhere along the dashed line and then get an expression for the electric eld a long distance away from the dipole
September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 2
Electric Field from an Electric Dipole
Start with two charges on the x-axis a distance d apart
Put -q at x = -d/2 Put +q at x = +d/2
Calculate the electric eld at a point P a distance x from the origin
September 8, 2011
Physics for Scientists & Engineers 2, Chapter 22
3
Electric Field from an Electric Dipole
e electric eld at any point x is the sum of the electric elds from +q and q
1 q 1 !q E = E+ + E! = + 2 4"# 0 r+ 4"# 0 r!2
Replacing r+ and r- we get the electric eld everywhere on the x-axis (except for x = d/2)
' q $ 1 1 E= & 2 # 2 ) 1 1 4!" 0 & ( x # 2 d ) ( x + 2 d ) ) % (
Interesting limit far away along the positive xaxis (x >> d)
September 8, 2011
Physics for Scientists & Engineers 2, Chapter 22
4
Electric Field from an Electric Dipole
We can rewrite our result as
q E= 4 0 x 2
=
-2 -2 d d 1 - - 1 + 2x 2x
We can use a binomial expansion or a Taylor expansion
d n , (1 ) 1 n + ... 2x d -2 (1 - ) = 1 - (-2) ... = 1 + 2 + ... 2x d -2 (1 + ) = 1 + (-2) ... = 1 - 2 + ... 2x
f (z) = (1 z)n f (z) =
n=0
f (n) (z = 0) (z - 0)n n!
For small z=d/2x, stop aer n=1
So we can write
E q 4 0 x 2 q 2d qd d d 1 + - 1 - = = x x 4 0 x 2 x 2 0 x 3
Physics for Scientists & Engineers 2, Chapter 22 5
September 8, 2011
Hint for Homework Problem
Exact dipole eld along x:
q Eex = 4 0 x 2
-2 -2 d d 1 - - 1 + 2x 2x
Approximated dipole eld along x for x>>d:
Eapprox qd = 2 0 x 3 Eex - Eapprox Eex Eapprox Eex
September 8, 2011
Relative error:
Eex - Eapprox Eex 0.05
0.95
Example: Relative error should be smaller than 0.05 solve the equation on the le for x/d
6
Physics for Scientists & Engineers 2, Chapter 22
Definition of Electric Dipole Moment
We de ne the electric dipole moment as a vector that points from the negative charge to the positive charge
! ! p = qd
p is the magnitude of the dipole moment q is the magnitude of one of the opposite charges d is the distance between the charges Using this de nition we can write the electric eld far away from an electric dipole as
p E= 2!" 0 x 3
September 8, 2011
Physics for Scientists & Engineers 2, Chapter 22
7
Electric Dipole Moment of Water
Demo: Polarization Chemistry reminder
the H2O molecule
e distribution of electric charge in an H2O molecule is non-uniform e more electronegative oxygen atom attracts electrons from the hydrogen atoms us, the oxygen atom acquires a partial negative charge and the hydrogen atoms acquire a partial positive charge e water molecule is "polarized"
September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 8
Electric Dipole Moment of Water
PROBLEM
Suppose we approximate the water molecules as two positive charges located at the center of the hydrogen atoms and two negative charges located at the center of the oxygen atom What is the electric dipole moment of a water molecule?
SOLUTION
e center of charge of the two positive charges is located exactly halfway between the centers of the hydrogen atoms e distance between the positive and negative charge centers is
September 8, 2011
d = r cos = (10-10 m ) cos(52.5) = 0.6 10-10 m 2
Physics for Scientists & Engineers 2, Chapter 22
9
Electric Dipole Moment of Water
is distance times the transferred charge, q = 2e, is the magnitude of the dipole moment of water
p = 2ed = 2 (1.6!10"19 C )( 0.6!10"10 m ) = 2!10"29 C m
is oversimpli ed calculation comes fairly close to the measured value of the electric dipole moment of water of 6.210-30 C m e fact that the real dipole moment of water is less than our calculated result indicates that the two electrons of the hydrogen atoms are not pulled all the way to the oxygen, but only one-third of the way
September 8, 2011
Physics for Scientists & Engineers 2, Chapter 22
10
Force Due to an Electric Field
e force exerted by an electric eld on a point charge is
! ! F = qE
e force vector is always tangent to the electric eld lines and points in the direction of the electric eld for a positive charge e force on a negative charge would be in the opposite direction
September 8, 2011
Physics for Scientists & Engineers 2, 22
11
Electric Chapter Dipole in Electric Field
Place a dipole in a uniform electric eld e dipole is composed of two charges, +q and q, separated by a distance d e net force on the dipole is zero e electric force on the two charges produces a torque given by
! = qEd sin"
Which can be written as e direction is given by the right-hand rule
September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 14
! ! ! ! = pE sin" or ! = p # E
General Charge Distributions
We have determine the electric eld from a single point charge and from several point charges Consider the electric eld due to a charge distribution We divide the charge into differential elements of charge, dq, and nd the electric eld from each differential charge element as if it were a point charge
dq = !dx $ & dq = " dA % for a charge distribution dq = #dV & ' dq dE = k 2 r (along a line & )over a surface & throughout a volume *
e magnitude of the electric eld is then
September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 17
Finite Line of Charge
To nd the electric eld along a line bisecting a nite length of wire with linear charge density , we integrate the contributions to the electric eld from all the charge We assume that the wire lies along the x-axis
September 8, 2011
Physics for Scientists & Engineers 2, Chapter 22
18
Finite Line of Charge
e symmetry of the situation allows us to conclude that there cannot be any electric force parallel to the wire We can calculate the electric eld due to all the charge for x 0 and multiply the result by 2 Consider a differential charge dq on the x-axis e magnitude of the the electric eld dE at a point (0,y) due to this charge is
dq dE = k 2 r r = x2 + y2
e component of the electric eld perpendicular to the wire is
dE y = k
September 8, 2011
dq dq " y % dqy cos! = k 2 $ ' = k 3 r2 r #r& r
Physics for Scientists & Engineers 2, Chapter 22 19
Finite Line of Charge
Taking dq = dx, the differential electric eld is
!dxy dE y = k 3 r
e electric eld at a distance y from the wire is then
"dxy E y = 2 ! dE y = 2 ! k 3 = 2k" y ! r 0 0 0
a a a
(
dx x +y
2 2
)
3
= 2k" y !
0
a
(x
dx
2
+y
2 3/2
)
e integral is
! (x
0
a
dx
2
+y
2 3/2
)
"1 =$ 2 $y #
% 1 ' = 2 2 2 x + y '0 y & x
a
a y 2 + a2
September 8, 2011
Physics for Scientists & Engineers 2, Chapter 22
20
Finite Line of Charge
e electric eld is then
" 1 E y = 2k! y $ 2 #y % 2k! '= 2 2 y y +a & a a y +a
2 2
a y 2 + a2
When the wire becomes in nitely long
a! " # !1
e electric eld a distance y from an in nitely long wire is
2k! Ey = y
September 8, 2011
Physics for Scientists & Engineers 2, Chapter 22
21
Ring of Charge
Consider a charged ring with radius R = 0.250 m e ring has uniform charge density and total charge of Q = +5.00 C What is the electric eld at d = 0.500 along the axis? e charge is evenly distributed around the ring We can calculate the electric eld by integrating the differential eld to do the differential charge By symmetry the eld will be parallel to the axis of the ring
September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 23
PROBLEM
SOLUTION THINK
Ring of Charge
We de ne an x-y coordinate system
SKETCH
RESEARCH
e differential electric eld, dE, at x = d is due a differential charge dq located at y = R e distance from (d,0) to (0,R) is e magnitude of the differential electric eld is
dq dE = k 2 r
September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 24
r = R2 + d 2
Ring of Charge
e magnitude of the x-component is
dEx = dE cos! = dE d r
e total electric eld is obtained by integrating dEx over all the charge in the ring
d dq Ex = ! dEx = ! k 2 r r ring ring
SIMPLIFY
We need to integrate around the circumference of the ring of charge using the differential arc length ds
Q dq = ds 2! R
September 8, 2011
Physics for Scientists & Engineers 2, Chapter 22
25
Ring of Charge
Now we write the integral
Ex =
2! R
(
0
" Q % d " kQd % k$ ds = # 2! R ' r3 $ 2! Rr3 ' & # &
2! R
(
0
kQd kQd ds = 3 = 2 2 3/2 r (R + d )
Putting in our numerical values
Ex
CALCULATE
(8.99!10 =
9
#( 0.250 m ) + ( 0.500 m ) % $ &
2 2
N m2 /C2 )(5.00!10"6 C )( 0.500 m )
= 128,654 N/C
We round our results to three signi cant gures
Ex = 1.29!105 N/C
ROUND
September 8, 2011
Physics for Scientists & Engineers 2, Chapter 22
26
Ring of Charge
Our units of N/C are correct Let's calculate the eld for d >> R
Ex =
DOUBLE-CHECK
(R
kQd +d
2
2 3/2
)
!
(d )
kQd
2 3/2
kQd kQ = 3 = 2 d d
Which is the result for a point charge, so our result seems reasonable Now calculate the eld for d = 0
Ex =
(R
kQd +d
2
2 3/2
)
=0
Which is what we would expect at the center of a ring of charge
September 8, 2011 Physics for Scientists & Engineers 2, Chapter 22 27