Problem #1
A charge of uniform linear density 2.0 nC/m is distributed along a long, thin, nonconducting rod.The rod is coaxial with a long conducting cylindrical shell (inner radius = 5.0 cm, outer radius = 10 cm). The net charge on the shell is zero. (a) What is the magnitude of the electric field 15 cm from the axis of the shell? What is the surface charge density on the (b) inner and (c) outer surface of the shell?
Answer:
Known:
Linear charge density on the thin rod λ = 2.0 × 10−9 C
Inner radius of the cylindrical shell R1= 5.0 × 10−2 m
Outer radius of the cylindrical shell R2 = 10 × 10−2 m
(a) Let us consider the portion of the cylinder of length L,
Total charge available on the thin rod for the length L is
q = Lλ = L x 2.0 x 10−9 C
So on the inner surface of the portion of the cylinder of length L this much charge q will be induced and the induced charge will have negative sign
So the charge induced
q = −L x 2.0 x 10−9 C
So surface charge density
σ1 = q/A = −L x 2.0 x 10−9/(2πR1)L
= −2.0 x 10−9 C/(2π x 5.0 x 10−2)
σ1 = −6.37 x 10−9 C/m2 = −6.37 nC/m2
(b) Charge induced on the outer surface of the cylinder for a length L ,
q = +L x 2 x 10−9 C
Surface charge density on the outer surface of the portion of the cylinder of length L is given by
σ2 = q/A = (+L x 2 x 10−9 C)/(2πR2)L
= (2.0 x 10−9 C)/(2π x 10 x 10−2 m)
σ2 = 3.18 × 10−9 C/m2 = 3.18 nC/m2
Problem #2
Figure is a section of a conducting rod of radius R1 = 1.30 mm and length L = 11.00 m inside a thin-walled coaxial conducting cylindrical shell of radius R2 = 10.0R1 and the (same) length L. The net charge on the rod is Q1 = +3.40 x 10─12 C; that on the shell is Q2 = ─2.00Q1. What are the (a) magnitude E and (b) direction (radially inward or outward) of the electric field at radial distance r = 2.00R2? What are (c) E and (d) the direction at r = 5.00R1? What is the charge on the (e) interior and (f) exterior surface of the shell?
Known:
Linear charge density on the thin rod λ = 2.0 × 10−9 C
Inner radius of the cylindrical shell R1= 5.0 × 10−2 m
Outer radius of the cylindrical shell R2 = 10 × 10−2 m
(a) Let us consider the portion of the cylinder of length L,
Total charge available on the thin rod for the length L is
q = Lλ = L x 2.0 x 10−9 C
So on the inner surface of the portion of the cylinder of length L this much charge q will be induced and the induced charge will have negative sign
So the charge induced
q = −L x 2.0 x 10−9 C
So surface charge density
σ1 = q/A = −L x 2.0 x 10−9/(2πR1)L
= −2.0 x 10−9 C/(2π x 5.0 x 10−2)
σ1 = −6.37 x 10−9 C/m2 = −6.37 nC/m2
(b) Charge induced on the outer surface of the cylinder for a length L ,
q = +L x 2 x 10−9 C
Surface charge density on the outer surface of the portion of the cylinder of length L is given by
σ2 = q/A = (+L x 2 x 10−9 C)/(2πR2)L
= (2.0 x 10−9 C)/(2π x 10 x 10−2 m)
σ2 = 3.18 × 10−9 C/m2 = 3.18 nC/m2
Problem #2
Figure is a section of a conducting rod of radius R1 = 1.30 mm and length L = 11.00 m inside a thin-walled coaxial conducting cylindrical shell of radius R2 = 10.0R1 and the (same) length L. The net charge on the rod is Q1 = +3.40 x 10─12 C; that on the shell is Q2 = ─2.00Q1. What are the (a) magnitude E and (b) direction (radially inward or outward) of the electric field at radial distance r = 2.00R2? What are (c) E and (d) the direction at r = 5.00R1? What is the charge on the (e) interior and (f) exterior surface of the shell?
Answer:
(a) magnitude E
we use
∮E.dA = qencl/ϵ0
For a cylinder:
∮E.dA = E(2πrL), then
qencl/ϵ0 = E(2πrL)
E = qencl/(2πrLϵ0)
E = λ/(2πrϵ0); WITH λ = qencl/L
(a) the magnitude E,
qencl = Q1 + Q2 , then
E = (Q1 + Q2)/(2πrLϵ0)
Because:
E = (─3.40 x 10─12 C)/(2π x 26 x 10─3 m x 11.00 m x 8.85 x10-12 C2/Nm2)
E = 0.214 N/C
(b) direction (radially inward or outward) of the electric field at radial distance r = 2.00R2 is radially inward since qencl < 0
(c) E for r = 5.00R1 = 5.00 x 1.3 mm = 6.5 x 10─3 m, and qencl = Q1 = +3.40 x 10─12 C, then
E = (+3.40 x 10─12 C)/(2π x 6.5 x 10─3 m x 11.00 m x 8.85 x10-12 C2/Nm2)
E = 0.855 N/C
(d) the direction at r = 5.00R1 is radially outward since qencl > 0
(e) the charge on the interior is
Qinner = ─ Q1 = ─3.40 x 10─12 C
(f) the charge on the exterior surface of the shell is
Qouter = Q2 ─ Qinner = (─2 x 3.40 x 10─12 C) ─ (─3.40 x 10─12 C)
Qouter = ─3.40 x 10─12 C
Problem #3
In Figure , short sections of two very long parallel lines of charge are shown, fixed in place, separated by L = 8.0 cm. The uniform linear charge densities are +6.0 μC/m for line 1 and ─2.0 μC/m for line 2. Where along the x axis shown is the net electric field from the two lines zero?
(a) magnitude E
we use
∮E.dA = qencl/ϵ0
For a cylinder:
∮E.dA = E(2πrL), then
qencl/ϵ0 = E(2πrL)
E = qencl/(2πrLϵ0)
E = λ/(2πrϵ0); WITH λ = qencl/L
(a) the magnitude E,
qencl = Q1 + Q2 , then
E = (Q1 + Q2)/(2πrLϵ0)
Because:
- Q1 + Q2 = +40 x 10─12 C +( ─2 x 3.40 x 10─12 C) = ─3.40 x 10─12 C,
- Radius, r = 00R2 = 2.00 (10R1) = 20 x 1.3 mm = 26 x 10─3 m
- length L = 00 m
E = (─3.40 x 10─12 C)/(2π x 26 x 10─3 m x 11.00 m x 8.85 x10-12 C2/Nm2)
E = 0.214 N/C
(b) direction (radially inward or outward) of the electric field at radial distance r = 2.00R2 is radially inward since qencl < 0
(c) E for r = 5.00R1 = 5.00 x 1.3 mm = 6.5 x 10─3 m, and qencl = Q1 = +3.40 x 10─12 C, then
E = (+3.40 x 10─12 C)/(2π x 6.5 x 10─3 m x 11.00 m x 8.85 x10-12 C2/Nm2)
E = 0.855 N/C
(d) the direction at r = 5.00R1 is radially outward since qencl > 0
(e) the charge on the interior is
Qinner = ─ Q1 = ─3.40 x 10─12 C
(f) the charge on the exterior surface of the shell is
Qouter = Q2 ─ Qinner = (─2 x 3.40 x 10─12 C) ─ (─3.40 x 10─12 C)
Qouter = ─3.40 x 10─12 C
Problem #3
In Figure , short sections of two very long parallel lines of charge are shown, fixed in place, separated by L = 8.0 cm. The uniform linear charge densities are +6.0 μC/m for line 1 and ─2.0 μC/m for line 2. Where along the x axis shown is the net electric field from the two lines zero?
Answer:
First you have to use Gauss's law to find electric field around the wire, then you use found formula to find place where the sum of two fields is equal to 0.
To find electric field around the wire select a cylindrical volume with axe coinsiding with wire. Let's this volume's height is h and radius is R. The total charge inside this volume is λh, where λ is linear charge density.
So the flux through the surface of your volume is
φ = λh/ϵ0.
From symmetry considerations lines of electric field are directed radially, so there is no flux through bases of the cylinder, only through side. From symmetry the electric field at the side of the cylinder is perpendicular to surface and is of the same value at every point on the side surface, so flux is simply the product of E by area of the surface that is A = 2πRh.
From Gauss's law we have
λh/ϵ0 = 2πRhE, so
E = λ/(2πϵ0R)
So electric field around the wire is inversely proportional to the first degree of distance R and directly proportional to linear charge density.
Now we can find position were fields from two wires are equal and of opposite direction. The requested point has to be to right of 2 or to the left of 1, closer to the weaker.
Because |λ2| < |λ1|the point must be to the right of wire 2, distance d from it.
First you have to use Gauss's law to find electric field around the wire, then you use found formula to find place where the sum of two fields is equal to 0.
To find electric field around the wire select a cylindrical volume with axe coinsiding with wire. Let's this volume's height is h and radius is R. The total charge inside this volume is λh, where λ is linear charge density.
So the flux through the surface of your volume is
φ = λh/ϵ0.
From symmetry considerations lines of electric field are directed radially, so there is no flux through bases of the cylinder, only through side. From symmetry the electric field at the side of the cylinder is perpendicular to surface and is of the same value at every point on the side surface, so flux is simply the product of E by area of the surface that is A = 2πRh.
From Gauss's law we have
λh/ϵ0 = 2πRhE, so
E = λ/(2πϵ0R)
So electric field around the wire is inversely proportional to the first degree of distance R and directly proportional to linear charge density.
Now we can find position were fields from two wires are equal and of opposite direction. The requested point has to be to right of 2 or to the left of 1, closer to the weaker.
Because |λ2| < |λ1|the point must be to the right of wire 2, distance d from it.
Field from first wire is
E1 = λ1/(2πϵ0R1) = (6 μC/m )/[2πϵ0(L + d)] = 6/(L + d)
and from second
E2 = λ2/(2πϵ0R2) = (─2 μC/m )/[2πϵ0(d)] = ─2/(d)
We have to find such x when
E1 + E2 = 0, then
{6/(L + d)]} + {─2/d} = 0
6/(L + d) = 2/(d)
we get,
d = L/2 = 4 cm to the right of wire 2, or
x = d + L = 12 cm to the left of wire 1
E1 = λ1/(2πϵ0R1) = (6 μC/m )/[2πϵ0(L + d)] = 6/(L + d)
and from second
E2 = λ2/(2πϵ0R2) = (─2 μC/m )/[2πϵ0(d)] = ─2/(d)
We have to find such x when
E1 + E2 = 0, then
{6/(L + d)]} + {─2/d} = 0
6/(L + d) = 2/(d)
we get,
d = L/2 = 4 cm to the right of wire 2, or
x = d + L = 12 cm to the left of wire 1
Post a Comment for "Applying Gauss’ Law: Cylindrical Symmetry Problems and Solutions2"