1. Dipole in a 2-D world because the 3-D world... (a) Find the electric field anywhere in the xy-plane (see...

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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).
Ans:
i
h
xı̂ı̂ı̂+(y+`)̂̂
xı̂ı̂ı̂+(y−`)̂̂
q
~
− 2
E = 4πo 2
2 3/2
2 3/2
[x +(y−`) ]
[x +(y+`) ]
(b) Evaluate the electric field found in part (a) on a circle with radius R (see Fig. 1).
Ans:
h
i
R cos θı̂ı̂ı̂+(R sin θ−`)̂̂
R cos θı̂ı̂ı̂+(R sin θ+`)̂̂
q
~
E = 4πo
− 2
2
2 3/2
2 3/2
[R −2`R sin θ+` ]
[R +2`R sin θ+` ]
(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 any terms containing square or higher powers
2
of R` because if R` is small, then R` is super-small.
Ans:
Ex = 4π3po R3 sin(θ) cos(θ), where p = 2`q
(d) Find the total charge enclosed by the circle. Find the unit-normal to the circle.
Ans: qenclosed = 0, n̂ = cos θı̂ı̂ + sin θ̂̂
PHGN200: All Sections
Recitation 1
January 23, 2007
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.
Ans:
hR
R 2π
2π
λ(xo −R cos θ)
R
~
ı̂
E = 4πo 0
dθ
−
3/2
0
(R2 +x2o −2xo R cos θ)
where λ = λo sin θ
λR sin θ
3/2 dθ
(R2 +x2o −2xo R cos θ)
i
̂
(b) Using only a symmetry argument, find the direction of the electric field.
Ans: in the negative y direction.
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).
Ans:
q
q
R x2
R x2
λ(a−x)
λ(b−f
(x))
2
1
1
~ =
E
1 + [f 0 (x)] dx ı̂ı̂+ 4πo x1
1 + [f 0 (x)]2 dx ̂
2
2 3/2
4πo x1 (a−x)2 +(b−f (x))2 3/2
[
]
[(a−x) +(b−f (x)) ]
Page 2
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