Download R r =Rdθ Q

Electricity and Magnetism: PHY-204
Fall Semester 2014
Recitation on Gauss’s Law
Solution
1. A sphere of radius R = 1.00 m surrounds a particle with charge Q = 50.0 µC located
at its center as shown in Fig. (1). Find the electric flux through a circular cap of
half-angle θ = 45.0◦ .
Fig. (1)
Answer
Using Gauss’s law, the electric flux through the circular cap can be find as
∫
⃗
⃗ · dA,
Φ=
E
A
⃗ the vector area element of circular cap of half angle θ is
where dA
dA = (circumference)(arc length)
= (2πr)dl = (2πR sin θ)Rdθ
from the geometry of figure,
=Rdθ
r
R
θ
dθ
Q
Date: 12 September, 2014
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Electricity and Magnetism: PHY-204
Fall Semester 2014
and the electric field due to a point charge Q located at center (or at a fixed distance
R) is
1 Q
r̂
4πε R2
)
∫ (0
)
1 Q (
2
⇒Φ =
2πR
sin
θ
dθ
4πε0 R2
∫ 45◦
Q
=
sin θdθ
2ε0 0
)
(
Q
Q
1
45◦
=
(− cos θ)|0 =
1− √
2ε0
2ε0
2
⃗ =
E
2. Find the electric field due to an infinitely long, thin wire of uniform linear charge
density λ. It will be useful to
(a) draw a sketch
(b) visualize the electric field based upon symmetry consideration.
Use Coulomb’s law.
Answer
(a)
p
y
θ
dx
O
x
(b) First we need to find the electric field due to a line element dx at point p, as
shown in figure, then integrate it from −∞ to +∞ to obtain the total field due
to an infinitely long wire.
The electric field due to one line element dx having a charge dq = λdx
1 dq
r̂
4πε0 r2
1
λdx
=
r̂
2
4πε0 (x + y 2 )
⃗ =
dE
Date: 12 September, 2014
from the geometry of figure.
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Electricity and Magnetism: PHY-204
Fall Semester 2014
Because of the symmetry of situation, the horizontal components of electric field
from two line elements lying at the same distance from origin, will cancel out. So
integrating over the vertical components only,
∫
∫
⃗
⃗
E = dE =
dE sin θĵ
∫
λ
ydx
=
ĵ,
2
4πε0
(x + y 2 )3/2
√
where we have substituted sin θ = y/ x2 + y 2 . Now integrate this integral over
the entire length of wire
∫ ∞
λy
dx
⃗
E =
ĵ
2
4πε0 −∞ (x + y 2 )3/2
∫ ∞
λ
dx
2
=
ĵ
by using
= ·
2
2
3/2
2πε0 y
y
−∞ (x + y )
3. Suppose Coulomb’s interaction is given not be an inverse square law, but by an 1/r3
dependence. Find the flux through a sphere of radius R centered on the point charge.
Answer
If the Coulomb’s interaction has an 1/r3 dependence then electric field due to a point
charge Q at distance r from the charge is
⃗ = C Q r̂
E
r3
where C is the constant of proportionality. So the flux through a sphere of radius R
centered on the point charge Q is
∫
∫
⃗
⃗
Φ=
E · dA =
A
2π
∫
π
C
0
0
∫
2π
∫
Q
R3
R2 sin θdθdϕ
|
{z
}
dA= vector area element
π
CQ
sin θdθdϕ
R 0
0
∫
CQ 2π
4πCQ
=
(2)dϕ =
·
R 0
R
=
So the electric flux has a 1/R dependence in this case which is against the empirical
observation.
Date: 12 September, 2014
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