Homework #2 Solutions Version 2
5 problems
Exercises
.
2
1. The figure shows an electron at the origin, and a grid marked off
in nanometers.
(a) what is the electric field (in component form) at the point
(1 nm, −2 nm), marked with a black star in the figure?
(b) If I put a proton at that point, what is the magnitude of the force
that it feels?
1
-2
e
-1
1
2
-1
-2
(1nm,–2nm)
/ (a) The electric field due to a point charge is given by the formula
~ = k qs ~r
E
r2 r
In this case, qs = −e = −1.6 × 10−19 C, so what we need to figure out is ~r. To get from the source to
the target, we go in the +x̂ direction a distance of 1 nm, and in the −ŷ direction a distance of 2 nm, so
~r = 1 nm(x̂) + 2 nm(−ŷ). The distance between source and target is then
p
√
r = |~r| = (1 nm)2 + (2 nm)2 = ( 5) nm = 2.24 × 10−9 m
and
~r
10−9 m(x̂) + (2 × 10−9 m)(−ŷ)
=
= 0.446x̂ − 0.893ŷ
r
2.24 × 10−9 m
(Notice that the components of unit vectors have no dimension: not meters or anything, just pure numbers.)
r̂ =
And so the electric field at the star is
~ = k qs r̂
E
2
r
2
−1.6 × 10−19 C
9 Nm
= 9 × 10
(0.446x̂ − 0.893ŷ)
C2
(2.24 × 10−9 m)2
= −2.87 × 108 N(0.446x̂ − 0.893ŷ)
= 1.28 × 108 N/C(−x̂) + 2.56 × 108 N/C(ŷ)
~ so the force on a proton (qt = 1.6×10−19 C)
(b) The force on a target charge qt in an electric field is F~ = qt E,
is simply
F~ = (1.6 × 10−19 C)(1.28 × 108 N/C(−x̂) + 2.56 × 108 N/C(ŷ))
= 2.05 × 10−11 N(−x̂) + 4.10 × 10−11 N(ŷ)
or
20.5 pN(−x̂) + 41.0 pN(ŷ)
(A piconewton is 10−12 N.) We note that this force points to the left (−x̂) and upwards (ŷ), which is back
towards the electron: exactly what we expect for oppositely charged objects.
The question asks for the magnitude of the force, so we use the Pythagorean theorem:
p
~
F = (20.5 pN)2 + (41.0 pN)2 = 45.8 pN
or 4.58 × 10−11 N .
2. Two positive charges, one with charge q1 = 2 µC and one with
charge q2 = 4 µC, sit on the y axis, 6 cm apart; the x axis runs right
between them. Find the electric field (magnitude and direction) on
the x axis, 4 cm to the right of the origin.
0.04m
0.03m
x
0.03m
q1
y
q2
/ The electric field due to two charges is the sum of the electric
field due to each charge:
~ =E
~1 + E
~ 2 = k q1 ~r1 + k q2 ~r2
E
r13
r23
and ~r2 = +(0.03 m)ŷ + (0.04 m)x̂
The
length of both vectors is the same:
p
(0.03 m)2 + (0.04 m)2 = 0.05 m. Now we solve
r1
=
r2
=
~ = k q1 ~r1 + k q2 ~r2
E
r13
r23
4 × 10−6 C
2 × 10−6 C
~r1 +
~r2
=k
(5 × 10−2 m)3
(5 × 10−2 m)3
2 × 10−6 C
= 9 × 109 N · m2 /C2
[~r1 + 2~r2 ]
125 × 10−6 m3
N
= 1.44 × 108
[−(0.03 m)ŷ + (0.04 m)x̂] + [(0.06 m)ŷ + (0.08 m)x̂]
C·m
N
8
= 1.44 × 10
(0.03 m ŷ + 0.12 m x̂)
C·m
= 17.3 MN/C x̂ + 4.3 MN/C ŷ = 1.73 × 107 N/C x̂ + 4.3 × 106 N/C ŷ
I asked for magnitude and direction, but that usually means I’m
really looking for component form. However, the direction is largely
to the right (since both charges are pushing to the right) and a little
bit up (because the lower charge, being twice as big, is pushing
harder). The magnitude is 1.8 × 107 N/C or 18 MN/C.
2
x
0.03m
where ~r1 is the vector from charge 1 to the target, and similarly
from ~r2 . We’re given q1 and q2 . From the diagram, we see that
~r1 = −(0.03 m)ŷ + (0.04 m)x̂
q1
y
0.03m
.
q2
r1
0.04m
r2
.
3.
In the figure, the four particles form a square of edge length
a = 5.00 cm and have charges q1 = +10.0 nC, q2 = −20.0 nC, q3 =
+20.0 nC, and q4 = −10.0 nC. In unit-vector notation, what net
electric field do the particles produce at the square’s center?
q1
q2
y
a
q4
x
q3
/ The electric field at the center of the square is the sum of the electric fields due to the four charges; and
as is the case with Coulomb’s Law, the “tricky” part is to find the vector ~r for each. For example, ~r1 is the
vector from q1 to the center, which can be gotten by moving a distance 21 a in the x̂ direction, and then 21 a
q
in the −ŷ direction; thus ~r1 = 12 ax̂ − 12 aŷ. The length of this vector is r1 = 14 a2 + 14 a2 = √12 a, and so its
unit vector is
1
a(x̂ − ŷ)
~r1
1
r̂1 =
= 2 1
= √ (x̂ − ŷ)
√ a
r1
2
2
By looking at the diagram, we see that the other three vectors all have the same length
and their vectors are
√1 a
2
(call this R)
1
1
1
a(−x̂) + a(−ŷ) and so r̂2 = √ (−x̂ − ŷ)
2
2
2
1
1
1
~r3 = a(−x̂) + a(ŷ) and so r̂3 = √ (−x̂ + ŷ)
2
2
2
1
1
1
~r4 = a(x̂) + a(ŷ) and so r̂4 = √ (x̂ + ŷ)
2
2
2
~r2 =
Thus the electric field at the center is
~ = k q1 r̂1 + k q2 r̂2 + k q3 r̂3 + k q4 r̂4
E
r12
r22
r32
r42
q1
q2
q3
q4
1
1
1
1
√ (x̂ − ŷ) + k
√ (−x̂ − ŷ) + k
√ (−x̂ + ŷ) + k
√ (x̂ + ŷ)
√
√
√
√
= k
2
2
2
2
(a/ 2)
2
(a/ 2)
2
(a/ 2)
2
(a/ 2)
2
k
√ [q1 (x̂ − ŷ) + q2 (−x̂ − ŷ) + q3 (−x̂ + ŷ) + q4 (x̂ + ŷ)]
=
(a2 /2) 2
√
k 2
=
[(q1 − q2 − q3 + q4 ) x̂ + (−q1 − q2 + q3 + q4 )ŷ]
a2
Now in this problem,
q1 − q2 − q3 + q4 = [(10) − (−20) − (20) + (−10)] nC = 0
−q1 − q2 + q3 + q4 = [−(10) − (−20) + (20) + (−10)] nC = 20 nC
Thus the electric field only has a y component: the x components of the fields here cancel. The field itself is
√
2
9 × 109 NCm2
2
~ =
E
(20 × 10−9 C)ŷ = 1.02 × 105 N/C ŷ or 102 kN/C
(0.05 m)2
3
Problems
Both of these problems require you to remember a bit from Physics 2130.
.
~ is the same at every point) exists in a region between oppositely charged
4. A uniform electric field (that is, E
plates. An electron is released from rest at the surface of the negatively charged plate and strikes the surface of
the opposite plate, 2.0 cm away, in a time 1.5 × 10−8 s.
(a) What is the speed of the electron as it strikes the second plate?
~
(b) What is the magnitude of the electric field E?
/ There is a uniform electric field between the plates, so the electron experiences a constant force, and thus
undergoes constant acceleration. Therefore we can use all of those wonderful constant-acceleration formulae
we learned in mechanics when dealing with gravity. Specifically,
1
x = x0 + v0 t + at2
2
and v = v0 + at
Solving the second equation for a gives us a = (v − v0 )/t; substituting into the first equation gives us
1
1
x − x0 = v0 t + (v − v0 )t = (v + v0 )t
2
2
which is a less familiar constant-acceleration formula (at least to me): the distance travelled is equal to the
time travelled times the average speed, which in the case of constant acceleration is (v + v0 )/2. The initial
velocity of the electron is v0 = 0, so the speed of the electron as it strikes the second plate is
v=2
∆x
(0.02 m)
=2
= 2.7 × 106 m/s
t
1.5 × 10−8 s
which answers part a. For part b, we need the electric field, which means we need the force, which means
we need the acceleration:
F
ma
E=
=
q
q
The acceleration comes from the equation
v = v0 + at
and so
a=
v
2.7 × 106 m/s
=
= 1.8 × 1014 m/s2 .
t
1.5 × 10−8 s
The charge of an electron is q = 1.6 × 10−19 C and the mass of an electron is m = 9.11 × 10−31 kg, so
E=
.
(9.11 × 10−31 kg)(1.8 × 1014 m/s2 )
= 1000 N/C .
1.6 × 10−19 C
5. In the figure, a small, nonconducting ball of mass m = 1.0 mg
(note the units) and charge q = 2.0 × 10−8 C (distributed uniformly
through its volume) hangs from an insulating thread that makes an
angle θ = 30◦ with a vertical sheet which has a uniform charge density
σ (shown in cross section); such a sheet creates a uniform electric field
which points away from or towards the sheet (depending on the sign),
σ
. Considering the gravitational force on the
with magnitude E =
20
ball and assuming the sheet extends far vertically and into and out of
the page, calculate the surface charge density σ of the sheet.
4
There are three forces on the ball, as shown in the figure to the
right: the tension T in the string, the weight mg of the ball, and
the electric force qE due to the field of the plate. The tension force
vector can be broken into its components T cos θ pointing upward
and T sin θ pointing to the left. Assuming the ball is not moving,
the net force on the ball is zero, and so
T sin θ = qE
and T cos θ = mg
/
We use the second equation to find T :
T =
mg
(1.0 × 10−6 kg)(9.8 m/s2 )
9.8 × 10−6 N
√
= 1.13 × 10−5 N
=
=
◦
cos θ
cos 30
3/2
Using this value, we solve for the electric field at the location of the ball:
E=
T sin θ
(1.13 × 10−5 N) sin 30◦
=
= 283 N/C
q
2.0 × 10−8 C
Now the electric field due to an infinite sheet of charge is E = 2πkσ. Therefore,
σ=
1
1
E=
(283 N/C) = 5.00 × 10−9 C/m2 = 5.00 nC/m2 .
2πk
2π(9 × 109 )
5