Quizz #2 solutions
PROBLEM 1
~ = −∇V = { 22ax2 2 , 22ay2 2 , 0}.
a. Electric field is E
(x +y )
(x +y )
b. The peculiarities are numerous. Some of them are: Ez = 0, the field does not depend on z, the magnitude of E
2
depends only on the distance from the z axis. To see that, let’s introduce rxy : x2 + y 2 = rxy
, so after setting
3
x = rxy cos θ and y = rxy sin θ, magnitude of the electric field is E = 2a/rxy . Here we can add some more
3
~ they are radial
characteristics: magnitude of E is proportional to 1/rxy
and if we look at the field lines of E,
straight rays lying in x-y plane, starting at the z axis. As magnitude of E depend on rxy only, lines of constant
magnitude of E as well as equipotential lines of V are circles.
PROBLEM 2
a. To calculate the electric field, let’s draw a cylinder around an original one, with length L2 << L and radius r.
Flux through the top and bottom flat parts of the cylinder is zero, and electric field is directed perpendicular
q
to the axis of the cylinder. Gaussian law gives us E · 2πrL2 = L2 ∗q/L
, so E = 2πr
.
0
0L
b. Let’s imagine the cylindrical sheet already has a charge q. Work done by the observer against the field to bring
additional charge dq from a to R can be calculated as:
Z R
Z a
Z a
Z a
Z a
~ =−
~ =
~ =
dW =
F~observer dr
F~observer dr
F~f ield dr
Ff ield dr =
(dq)Edr.
(1)
a
R
Taking integral of E, we get dW =
charge the shell:
Z
W =
R
(dq) 2πq0 L
Q
0
R
R
a
· ln R
. Now we want to calculate total work needed to gradually
Z
dW =
0
Q
a
=
(dq) 2πq0 L · ln R
You may find it familiar as it resembles formula for capacitors U =
Q2
1
2 2π0 L
a
· ln R
.
(2)
2
1Q
2 C .
c. Total work done on charges would be a change of a sum of kinetic energies of the charges, which is zero. So, the
2
a
.
work done by electric field is opposite to the above, Wf ield = − Q2 2π10 L · ln R
d. While charging, change in the potential energy is given by ∆U = −Wf ield = Wobserver . At the beginning, it’s
2
a
natural to set potential energy to zero, so the final U = Q2 2π10 L · ln R
.
PROBLEM 3
a. We draw a sphere with the radius r < R0 as a Gaussian surface and apply Gauss’s law:
E · 4πr2 =
so E =
Qinside
0
=
4 3
3 πr ρ ,
0
(3)
rρ
30 .
b. For R1 > r > R0 , the Gaussian surface will change accordingly and will be a sphere with radius r between
conducting and nonconducting parts. The right-hand side of the equation changes to be total charge of the
nonconductive sphere of radius R0 over 0 : E · 4πr2 =
Qinside
0
=
4
3
3 πR0 ρ ,
0
so so E =
R03 ρ
3r 2 0 .
c. Inside conductor elector field is zero, so for R2 > r > R1 E = 0.
d. For r > R2 , the right-hand side of the Gauss’s law will be the total charge of the system over 0 : E · 4πr2 =
Qinside
0
=
4
3
3 πR0 ρ+Q ,
0
so E =
4
3
3 πR0 ρ+Q .
4πr 2 0
2
PROBLEM 4
The thickness d of the slab is small comparing to the breadth and height of the slab. As we are interested only
in the field close to the slab (distances of order d), we can treat the slab as a set of infinite planes and assume the
electric field is always perpendicular to the slab surface (parallel or antiparallel to x̂).
a. Our Gaussian surface will be a cylinder oriented along x̂ and symmetrical relative to the center of the slab: it
has a flat circular surfaces of area A at coordinates +x and −x. The Gauss’s theorem tells us that E · 2A =
xρ
Qinside
= ρA·2x
0
0 . So, electric field inside is E = 0 .
b. Same Gaussian surface, but now flat circular surfaces of the cylinder are located outside of the cylinder, so a
formula for total charge inside the cylinder changes: E · 2A = Qinside
= ρA·d
0
0 . So, electric field outside close to
dρ
the surface is E = 20 .