Physics 2220 – Module 03 Homework
01.
What is the electric potential energy of the group of charges in the figure?
The electric potential energy will be the sum of the potential
energy of each pair of charges. In this case:
U = ke
9
(
2
q 1 q 2 q 1 q3 q2 q 3
+
+
r 12
r 13
r 23
2
U = (8.99 × 10 N m / C )
)
(−3.0 nC) (2.0 nC) (−3.0 nC) (−3.0 nC) (−3.0 nC)
+
+
( (2.0 nC)
)
(0.04 m)
0.04 m
0.04 m
U = 6.7 × 10−7 J
02.
What potential difference is needed to accelerate the He + ion (charge +e, mass 4u) from rest to a speed
of 3.00 ⨯ 106 m/s?
The He+ ion has one missing electron, so it will have a net charge of:
−19
q = + 1.60 × 10
C
The mass can also be found with the given information:
−27
m = 4u = 4 (1.67 × 10
−27
kg) = 6.68 × 10
kg
Assume that conservation of mechanical energy applies.
E f = Ei
K f + U f = Ki + Ui
Kf − Ki = Ui – U f
→
Δ K = −Δ U
Recall the relationship between electric potential and electric potential energy:
Δ U
q
−Δ K
Δ V=
q
Δ V=
The mass starts from rest and accelerates to a final speed, so the initial velocity is zero:
Δ V=
−mv 2f
2q
Solve for the needed potential difference:
Δ V=
−(6.68 × 10−27 kg) (3.00 × 10 6 m/s)2
2 (1.60 × 10−19 C)
Δ V = −1.88 × 10 5 V
03.
(a)
What is the potential of an ordinary AA or AAA battery?
Electric potential of a ordinary AA or AAA battery is indicated on the battery:
V = 1.5 V
(b)
An AA battery is connected to a parallel-plate capacitor having 4.0 cm ⨯ 4.0 cm plates spaced
1.0 mm apart. How much charge does the battery supply to each plate?
Each plate on the capacitor will have a surface charge density which is related to the total
charge:
σ=
Q
Q
=
A
xy
→
Q = σ xy
The electric field above the surface charge has been shown to be:
E = ϵσ0
→
Q = E ϵ0 xy
Finally, the relation between the potential difference and the electric field for a parallel-plate
capacitor has been shown to be:
Δ V = Ed
→
Q=
Δ V ϵ0 xy
d
Solve for the charge:
Δ V ϵ0 xy
d
−12
2
2
(1.5 V) (8.85 × 10 C / N m ) (0.04 m) (0.04 m)
Q=
= 2.1 × 10−11 C
(0.001 m)
Q=
04.
In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm.
(a)
What is the electric potential of the proton at the position of the electron?
Treat the nucleus of the hydrogen atom as a point charge:
q
r
(1.60 × 10−19 C)
9
2
2
V = (8.99 × 10 N m / C )
= 27 V
(0.053 × 10−9 m)
V = ke
(b)
What is the electron's potential energy?
Use the relation between electric potential and electric potential energy:
−19
U = (−1.6 × 10
U = qV
C) (27 V) = −4.3 × 10− 18 J
05.
The electric potential at the dot in the figure is 3140 V.
What is charge q?
The total potential at the dot is the sum of the potential each charge
contributes to the total.
V = ke
(
q 1 q 2 q3
+
+
r1
r2
r3
)
Where it is assumed:
q1 = 5.0 nC
r 1 = 4.0 cm
q 2 = −5.0 nC
r 2 = 2.0 cm
r 3 = √ r 21 + r 22 = √(4.0 cm)2 + (2.0 cm)2 = 4.5 cm
Solve for q3.
q3 = r 3
q 3 = (0.045 m)
(
(
q
q
V
− 1− 2
ke
r1 r 2
)
3140 V
5.0 × 10−9 C −5.0 × 10−9 C
−
−
0.040 m
0.020 m
8.99 × 10 9 N m 2 / C2
)
q3 = 2.13 × 10−8 C
06.
A -10.0 nC point charge and a +20.0 nC point charge are 15.0 cm apart on the x-axis.
(a)
What is the electric potential at the point on the x-axis where the electric field is zero?
Place the 20.0 nC charge at the origin and the -10.0 nC 15 cm to the right. The distance from
the origin to the point on the x-axis where the electric field is zero is x, and the distance from the
-10.0 nC charge will then be x – 15.0 cm. This is because the field vectors need to cancel out to
zero (see above).
⃗
E = E⃗1 + E⃗2 = 0
The electric field point and charges are all along the x-axis, so the vector notation can be
dropped and distance determined.
E 1 + E2 = k e
q1
q1
x
2
+ ke
q2
(x − 15.0 cm)2
q2
=0
=−
x2
(x − 0.15 m)2
+ 20.0 nC
−10.0 nC
2
1
=−
→
=
2
2
2
x
(x − 0.15 m)
x
(x − 0.15 m)2
x 2 − 60.0 cm x + 450 cm 2 = 0
At this point, use the quadratic formula to solve for x
x = 51.2 cm
and
x = 8.8 cm
x = 8.8 cm is not possible because the electric field vectors from both charges will be pointing to
the right. Therefore x = 51.2 cm.
Now solve for the total potential at this position created by both charges:
V = V 1 + V 2 = ke
9
2
2
V = (8.99 × 10 N m / C )
(b)
q
( qx + x − 0.15
m)
1
2
20.0 nC
−10.0 nC
+
= 103 Volts
( 0.512
m 0.512 m − 0.15 m )
What is the magnitude of the electric field at the point on the x-axis, between the charges, where
the electric potential is zero?
The electric potential contributions from each charge will add up to zero:
q1
q2
+
=0
x
15.0 cm − x
q1
q2
=−
x
15.0 cm − x
q 1 (15.0 cm − x ) = −q 2 x
→
q 1 (15.0 cm) = x (q 1 − q 2 )
q1
20.0 nC
x = (15.0 cm)
= (15.0 cm)
= 10.0 cm
q1 − q2
20.0 nC − (−10.0 nC)
V = V1 + V 2 = ke
(
)
Now find the electric field. REMEMBER this is a point between the two particles, so the electric
field for each charge will point to the right. Use magnitude of the charges.
⃗ = E⃗1 + E⃗2 = k e
E
⃗
E = (8.99 × 10 9 N m2 / C2 )
07.
(
(
q1
x
2
+
q2
(15.0 cm − x )2
)
̂i
20.0 nC
∣−10.0 nC∣ ̂
+
i = 5.4 × 10 4 N/C ̂i
2
(0.100 m)
(0.050 m)2
)
What is the escape speed of an electron launched from the surface of a 1.0-cm-diameter glass sphere
that has been charged to 10 nC?
Use conservation of energy with all the assumptions that are needed for
the escape velocity.
vf = 0
rf = ∞
→
→
Kf = 0
Uf = 0
The electron must travel through potential from the surface of the glass sphere to infinity. Solve for the
escape speed (initial velocity).
Kf + U f = K i + Ui
1
0 + 0 = me v 2i + q e V
2
1
Q
0 = me v 2i + q e k e
2
R
( )
√
√
−2 q e k e Q
−2 (−1.60 × 10−19 C) (8.99 × 10 9 N m2 / C 2) (10 × 10−9 C)
vi =
=
me R
(9.11 × 10−31 kg ) (0.005 m)
v i = 8.0 × 107 m/s