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PHYSICS 120 : ELECTRICITY AND MAGNETISM
TUTORIAL QUESTIONS
GAUSS’ LAW
Question 43
A thin-walled metal sphere has a radius of 25.0 cm and carries a charge of 2.00 × 10−7 C.
Find E for a point (a) inside the sphere, (b) just outside the sphere, and (c) 3.00 m from the
centre of the sphere.
(a) Inside the shell there is no charge, so
ψ=
QENC
=0
ǫ0
hence the field inside the shell is zero.
(b) To find the field just outside the shell, choose the Gaussian surface to be a sphere, with
surface area 4πr2 , so using Gauss’ Law:
QENC
ψ=
= E × (4πr2 )
ǫ0
1 QENC
∴E =
4πǫ0 r2
2 × 10−7
= (9 × 109 )
6.25 × 10−2
= 2.88 × 104 N C−1
(c) At a distance r = 3.00 m away:
1 QENC
4πǫ0 r2
2 × 10−7
= (9 × 109 )
3.02
−1
= 200 N C
E =
Question 44
Two charged concentric spheres have radii of 10.0 cm and 15.0 cm. The charge on the inner
sphere is 4.00×10−8 C and that on the outer sphere −2.00×10−8 C. Find E at (a) r = 12.0 cm
and (b) r = 20.0 cm.
page 1 of 5
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
(a) At a distance r = 12 cm, the field due to the outer sphere is zero, hence the enclosed
charge is the charge of the inner sphere: QENC = QI . The electric field is then
QENC
ǫ0 A
1
4 × 10−8
=
4πǫ0 (12 × 10−2 )2
= 2.50 × 104 N C−1
E =
(b) Let EI and EO be the fields due to the inner and outer spheres. The fields are
EI =
1 QI
4πǫ0 r2
4 × 10−8
(20 × 10−2 )2
= 9.00 × 103 N C−1 outwards
−2 × 10−8
= (9 × 109 )
(20 × 10−2 )2
= −4.50 × 103 N C−1 inwards
= (9 × 109 )
EO
By the principle of superposition, the resultant field is
E = EI + EO = 4.50 × 103 N C−1 radially outwards
page 2 of 5
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 45
A small sphere whose mass m is 1 mg carries a charge q of 0.02 µC. It hangs from a silk thread
which makes an angle of 30° with a large, charged conducting sheet as shown. Calculate the
charge density σ for the sheet.
Since the system is in equilibrium, the forces are balanced as so Ty = W and Tx = F , where
T is the tension in the string, W is the weight of the sphere and F is the repulsive force
between the sphere and the sheet.
Since the electric field due to the sheet is
E=
σ
2ǫ0
whereσ is the surface charge density. The force F is then given by
E=
σ
F
=
q
2ǫ0
The x- and y-components of the tension are
1
Tx = T sin 30° = T
2√
3T
Ty = T cos 30° =
2
√
1
3T
and F = T
∴W =
2
2
2
⇒ 2F = √ W
3
1
F = √ mg
3
The surface charge density σ is then
2ǫ0 1
√ mg
q 3
(2) × (8.85 × 10−12 ) × (1 × 10−6 ) × (9.8)
√
=
3 × (0.02 × 10−6 )
= 5.01 × 10−9 C m−2
σ =
page 3 of 5
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 46
Two uniformly charged planes with surface charge densities of 3.00×10−9 C m−2 and −1.00×
10−9 C m−2 lie parallel to each other a distance 8.00 cm apart in a vacuum. Draw the field
lines between the plates and behind each plate and find E everywhere. (Hint: use superposition.)
Assume infinite sheets (so, therefore, no edge effects) and use the result E = σ/(2ǫ0 ) for a
plane surface with surface charge density σ.
For E1 , the electric field depends on the electric field EA due to sheet A and the electric
field EB due to B. The magnitude of E1 is
|E1 | = ||EA | − |EB ||
1
=
||σA | − |σB ||
2ǫ0
1
=
× ((3 − 1) × 10−9 )
−12
8.85 × 10
= 113 N C−1
2π 1
||σA | − |σB || in terms of the Coulomb constant ke
=
2π 2ǫ0
2π
=
||σA | − |σB ||
4πǫ0
1
= (2π)
||σA | − |σB ||
4πǫ0
= (2π) × (9 × 109 ) × (2 × 10−9 )
= 36π N C−1
Let E2 be the electric field between the plates A and B. In this region, the electric field EA
due to A points in the same direction as the electric field EB due to B. Therefore
|E2 | = ||EA | + |EB ||
1
||σA | + |σB ||
= (2π)
4πǫ0
= (2π) × (9 × 109 ) × ((3 + 1) × 10−9 )
= 72.0π N C−1
Similarly, it can be shown that for E3
|E3 | = |−|EA | + |EB || = |E1 |
page 4 of 5
PHYS120 ELECTRICITY AND MAGNETISM: TUTORIAL QUESTIONS
Question 47
An infinite charged sheet has a charge density σ of 0.10 nC mm−2 . How far apart are the
equipotential surfaces whose potential differ by 5.00 kV?
The field due to an infinite charged sheet is
E=
σ
2ǫ0
The electric potential is given by the field E and the separation d of the plates:
V = Ed
For equipotential surfaces whose potential differ by 5.00 kV
∆V = |V2 − V1 | = 5.00 × 103 = d ×
Therefore the separation d is
2ǫ0
× (5.00 × 103 )
σ
4πǫ0
5.00 × 103
=
×
2π
0.1 × 10−9 × 106
= 8.8 × 10−4 m
d =
page 5 of 5
σ
2ǫ0