PHGN200: All Sections
Recitation 1
January 23, 2007
y
R
q
`
θ
x
`
−q
Figure 1: For the above electric dipole, p = 2`q.
1. Dipole in a 2-D world because the 3-D world is too damn hard!
(a) Find the electric field anywhere in the xy-plane (see Fig. 1).
Solution: The electric field due to N point particles is given by
~ =
E
N
1 X ~ri
qi ,
4πo i=1 ri 3
where ~r = ~rf − ~rsi .
(1)
point.
~rf is called the field point and ~rsi is the ith source point
We will call the positive charge the first charge and the negative charge the second charge. Thus,
~rf = xı̂ı̂ + y̂̂
~rs1 = `̂̂
~rs2 = −`̂̂̂,
Computing all quantities needed for (1) yields
~r1 = ~rf − ~rs1
= xı̂ı̂ + (y − `)̂̂
r1 = ||~rf − ~rs1 ||
p
= x2 + (y − `)2
~r2 = ~rf − ~rs2
= xı̂ı̂ + (y + `)̂̂
r2 = ||~rf − ~rs2 ||
p
= x2 + (y + `)2 .
Finally, substituting (2) and (3) into (1) yields
#
"
ı̂
̂
ı̂
̂
q
xı̂
+
(y
−
`)̂
xı̂
+
(y
+
`)̂
~ =
−
E
4πo [x2 + (y − `)2 ]3/2 [x2 + (y + `)2 ]3/2
(2)
(3)
(4)
PHGN200: All Sections
Recitation 1
January 23, 2007
(b) Evaluate the electric field found in part (a) on a circle with radius R (see Fig. 1).
Solution: To find the electric field on the circle, we set the field point on the circle, i.e., x = R cos θ and
y = R sin θ in (4), which yields
"
#
ı̂
̂
ı̂
̂
q
R
cos
θı̂
+
(R
sin
θ
−
`)̂
R
cos
θı̂
+
(R
sin
θ
+
`)̂
~ =
E
−
(5)
4πo [R2 − 2`R sin θ + `2 ]3/2
[R2 + 2`R sin θ + `2 ]3/2
(c) Find an approximate expression for the x-component of the electric field on the circle if `/R 1 (see
Fig. 1). Hint: Expand the denominator in Taylor series, (1 + )n = 1 + n + · · · , if || < 1. You can drop
2
any terms containing square or higher powers of R` because if R` is small, then R` is super-small.
Solution: From (5), we see that the x-component of the electric field on the circle is given by
"
#
qR cos θ
1
1
−
.
Ex =
4πo
[R2 − 2`R sin θ + `2 ]3/2 [R2 + 2`R sin θ + `2 ]3/2
We rewrite the denominators as follows:
1
[R2 ∓ 2`R sin θ +
`2 ]3/2
−3/2
= R2 ∓ 2`R sin θ + `2
2 !#−3/2
2`
sin
θ
`
= R2 1 ∓
+
R
R
"
2 #−3/2
2`
sin
θ
`
= R−3 1 ∓
+
R
R
"
2` sin θ
+
= R [1 + ]
, where = ∓
R
3
= R−3 1 − + · · ·
2
"
#
2
3`
sin
θ
3
`
= R−3 1 ±
−
+ ···
R
2 R
3` sin θ
−3
≈R
1±
.
R
−3
−3/2
Finally, substituting the above result into (6) and simplifying yields
Ex =
3p
sin(θ) cos(θ),
4πo R3
Page 2
where p = 2`q.
2
`
R
(6)
PHGN200: All Sections
Recitation 1
January 23, 2007
(d) Find the total charge enclosed by the circle. Find the unit-normal to the circle.
Solution: The total charge enclosed by the circle is given by
qenclosed = q1 + q2
= q + (−q)
=0
We will find the radially-outward unit-normal for the upper-half of the circle (you should do both!). The
equation for the circle (upper-half) with radius R is given by
p
R = x2 + y 2 .
The non-unit normal is given by (review Calculus!)
∂R
∂R
ı̂ +
̂
∂x
∂y
1
=p
(xı̂ı̂ + y̂̂̂) .
2
x + y2
n̂ =
(7)
Substituting x = R cos θ and y = R sin θ into (7) yields
~n = cos (θ)ı̂ı̂ + sin (θ) ̂̂.
(8)
It is trivial to confirm that ~n given in (8) is actually n̂ (we got lucky!).
2. Hard integrals made easy!
(a) Set-up an integral expression for the electric field at a field point, (xo , 0), due to a ring of charge with
linear charge density λ = λo sin θ, where λo is a positive constant (see Fig. 2).
y
R
θ
(xo, 0)
x
Figure 2: The circle has a linear charge density given by λ = λo sin θ, where λo is a positive constant
constant.
Page 3
PHGN200: All Sections
Recitation 1
January 23, 2007
Solution: The electric field is given by
~ =
E
1
4πo
Z
~r
dq,
r3
(9)
where ~r = ~rf − ~rs . From Fig. 2, we have
~rf = xoı̂ + 0̂̂
~rs = R cos(θ)ı̂ı̂ + R sin(θ)̂̂̂.
Thus,
~r = (xo − R cos θ)ı̂ı̂ − R sin θ̂̂
p
r = R2 + x2o − 2xo R cos θ.
(11)
dq = λdrs ,
(12)
(10)
dq is given by
where drs is the differential element of the source variable (we are summing up the source points, NOT the
field points, hence, we use rs NOT rf ). Also, note that dq is a scalar, thus, the right hand side of (12) must
also be a scalar. One way to find drs is as follows:
~rs
d~rs
dθ
drs
dθ
drs
dθ
drs
dθ
drs
= R cos(θ)ı̂ı̂ + R sin(θ)̂̂
= −R sin(θ)ı̂ı̂ + R cos(θ)̂̂
d~rs = dθ q
= R2 cos2 θ + sin2 θ
=R
= Rdθ.
Finally, substituting (10), (11), and (13) into (9) yields
"Z
#
Z 2π
2π
λ
(x
−
R
cos
θ)
λR
sin
θ
R
o
~ =
E
dθ ı̂ −
dθ ̂ ,
4πo 0 (R2 + x2o − 2xo R cos θ)3/2
(R2 + x2o − 2xo R cos θ)3/2
0
where λ = λo sin θ.
(b) Using only a symmetry argument, find the direction of the electric field.
Page 4
(13)
(14)
PHGN200: All Sections
Recitation 1
January 23, 2007
Solution:
y
R
θ
(xo, 0)
x
Green = Blue + Red
Figure 3: We know that the electric field points radially outward from a positive charge (shown in red) and
radially inward from a negative charge (shown in blue). The linear charge density is given by λ = λo sin θ,
thus, the upper half of the circle is positively charged and the lower half of the circle is negatively charged.
By drawing a few “representative” charges on the circle, we readily see that the electric field has ONLY a
y−component. Moreover, we see that the electric field points down, i.e., in the negative y direction. It is
interesting to note that we have effectively calculated the x−component of the integral given by (14) just by
drawing a picture!
3. Given a curve y = f (x) where x1 ≤ x ≤ x2 in the xy−plane, find the electric field at some field point, (a, b).
Assume that the curve has a linear charge density given by λ = g(x).
Page 5
PHGN200: All Sections
Recitation 1
January 23, 2007
Solution: The electric field is given by
~ =
E
1
4πo
Z
~r
dq,
r3
(15)
where ~r = ~rf − ~rs . First we find ~rf and ~rs as follows:
~rf = aı̂ı̂ + b̂̂
~rs = xı̂ı̂ + y̂̂
=xı̂ı̂ + f (x)̂̂ recall that y = f (x)
Thus,
~r = [a − x]ı̂ı̂ + [b − f (x)] ̂
q
r = [a − x]2 + [b − f (x)]2 .
(16)
(17)
dq is given by dq = λdrs and can be found in the usual manner,
~rs = xı̂ı̂ + f (x)̂̂
d~rs
= 1ı̂ı̂ + f 0 (x)̂̂
dx
drs d~rs = dx
dx q
= 1 + [f 0 (x)]2
q
drs = 1 + [f 0 (x)]2 dx.
(18)
Note that (18) is just an arc length formula from Calculus. Finally, substituting (16), (17) and (18) into (15)
yields
Z x2
q
1
λ(a − x)
~
E=
1 + [f 0 (x)]2 dx ı̂
4πo x1 (a − x)2 + (b − f (x))2 3/2
Z x2
(19)
q
1
λ(b − f (x))
2
0
̂
+
1 + [f (x)] dx ̂,
4πo x1 (a − x)2 + (b − f (x))2 3/2
where λ = g(x).
Page 6