Problem #1
An electron is released 10.0 cm from a very long nonconducting rod with a uniform 5.0 μC/m. What is the magnitude of the electron’s initial acceleration?Answer:
Known:
distance r = 10.0 cm = 0.1 m
Linear charge density λ = 5.0 X 10-6 C/m
Electric field at distance r from the non conducting rod
E = λ/(2πε0r)
where ε0 = permitivity of free space = 8.85 x10-12 C2/Nm2
E = (5.0 X 10-6 C/m)/(2π x 8.85 x10-12 C2/Nm2 x 0.1)
E = 8.99 x 105 N/C
the magnitude of the electron's initial acceleration given by
E = F/q = ma/q
where q = charge of electron = 1.6 x 10-19 C
m = mass = 9.11 x 10-31 kg
a = Eq/m = (8.99 x 105 N/C)(1.6 x 10-19 C)/(9.11 x 10-31 kg )
a = 1.58 x 1017 m/s2
Problem #2
(a) The drum of a photocopying machine has a length of 42 cm and a diameter of 12 cm. The electric field just above the drum’s surface is 2.3 x 105 N/C. What is the total charge on the drum? (b) The manufacturer wishes to produce a desktop version of the machine. This requires reducing the drum length to 28 cm and the diameter to 8.0 cm.The electric field at the drum surface must not change. What must be the charge on this new drum?
Answer:
(a) Known: length of l = 42 cm = 0.42 m,
diameter of, d = 12 cm = 0.12 m
electric field just above the drum’s surface is E = 2.3 x 105 N/C
then, total charge on the drum given by is
q = σA = σ(πdh)
q = πdhEϵ0
= π x 0.12 m x 0.42 m x 2.3 x 105 N/C x 8.85 x10-12 C2/Nm2
q = 3.2 x 10─7 C = 0.32 μC
(b) New area A = πd’h’, then
q' = q(A’/A) = (d’h’)/(dh)
= (3.2 x 10─7 C)(8.0 cm x 28 cm)/(12 cm x 42 cm)
q’ = 1.4 x 10─7 C = 0.14 μC
Problem #3
Figure shows a section of a long, thin-walled metal tube of radius R = 3.00 cm, with a charge per unit length of λ = 2.00 x 10─8 C/m. What is the magnitude E of the electric field at radial distance (a) r = R/2.00 and (b) r = 2.00R? (c) Graph E versus r for the range r = 0 to 2.00R.
Known:
a charge per unit length of λ = 2.00 x 10─8 C/m
thin-walled metal tube of radius R = 3.00 cm
- E must Be RADIAL BY Symmetry
- qenclosed = λΔZ with λ= σ x 2Πr
(a) For r < R, qenclosed = 0, then so r = R/2.00 E = 0.
(b) For r > R
E.A = E x 2πRΔZ = λΔZ/ϵ0, then
E = λ/2πRϵ0
= (2.00 x 10─8 C/m)/(2π x 0.03 m x 8.85 x10-12 C2/Nm2)
E = 1.19 x 104 N/C
(c) Graph E versus r for the range r = 0 to 2.00R.
Problem #3
An infinite line of charge produces a field of magnitude 4.5 x 104 N/C at distance 2.0 m. Find the linear charge density.
Answer:
Known:
field of magnitude E = 4.5 x 104 N/C
distance d = 2.0 m
Electric field roduced by the infinite line at a distance d having linear change density λ is given by the relation,
E = λ/2πRϵ0
4.5 x 104 N/C = λ/(2π x 2.0 m x 8.85 x10-12 C2/Nm2)
λ = 1.0 x 10─5 C/m
Therefore, the linear charge density is 1.0 x 10─5 C/m
Problem #4
A long straight wire has fixed negative charge with a linear charge density of magnitude 7.6 nC/m. The wire is to be enclosed by a coaxial, thin-walled nonconducting cylindrical shell of radius 1.5 cm. The shell is to have positive charge on its outside surface with a surface charge density s that makes the net external electric field zero. Calculate σ.
Answer:
Known:
linear charge density of magnitude λ = 7.6 nC/m = 7.6 x 10─9 C/m
thin-walled nonconducting cylindrical shell of radius, R = 1.5 cm = 1.5 x 10─2 m
the total electric field on the cylinder and wire is
E = Ewire + Ecylinder
E = (─λ/2πrϵ0) + (λ’/2πrϵ0)
Net charge charge cylindrical is
q = λ’L (*)
or net charge cylindrical shell is
q = σA = σ(2πRL) (**)
from (*) and (**) we get
λ’L = σ(2πRL)
λ’ = 2πRσ
Net charge enclosed should be zero since net flux (net electric Field) is zero
0 = (─λ/2πrϵ0) + (λ’/2πrϵ0)
λ = 2πRσ
σ = λ/2πR
= (7.6 x 10─9 C/m)/(2π x 1.5 x 10─2 m)
σ = 8.06 x 10─8 C/m2 = 80.6 nC/m2
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