# Chapter 16 Problem 1 ```Chapter 16
Problem 1
The magnitude of the work is given by the product |W | = qE∆x cos θ = 6.4 &middot; 10−19 J. The angle is zero
because the motion of the proton is parallel to the field. The proton moving along a field loses potential
energy, thus the change in potential energy is also 6.4 &middot; 10−19 J. The potential difference is V = E∆x = 4V
Problem 3
The work required is going to be W = qE∆x = q∆V . Here, a sodium ion has a total charge of +1e =
1.6 &middot; 10−19 C, and ∆V = 90mV . Therefore the work required is 90mV &middot; 1.6 &middot; 10−19 C = 1.44 &middot; 10−20 J.
Problem 11
The electric potential from a single charge is given by V = kre q where r is the distance from the charge.
Plugging in r = 0.01m gives V = 1.44 &middot; 10−7 V . The potential difference between two points, ∆V = V2 − V1 ,
is found by evaluating hte potential at each of the two points, and subtracting them. The first point has
already been evaluated, the potential at the second point is 7.2 &middot; 10−8 V , thus the difference is −7.2 &middot; 10−8 V ,
negative because V1 is bigger than V2 .
Problem 13
The potential is found by adding the potential due to each of the charges. Note that there are no VECTORS
involved here, just numbers:
P ke q
9
9
9&middot;109 ∗8&middot;10−6
∗4&middot;10−6
∗2&middot;10−6
+ 9&middot;10 0.03
+ 9&middot;100.0671
= 2.67 &middot; 106 V . If we replace the bottom left charge
V =
r =
0.06
with −2&micro;C, then the third term in the above sum becomes negative. This changes the answer to 2.13 &middot; 106 V .
Problem 15
Once again, add the potential due to each charge separately. Note that Q2 is negative. The potential half
way between (r = 0.175m):
P ke q
9
9&middot;109 ∗5&middot;10−9
∗−3&middot;10−9
V =
+ 9&middot;10 0.175
= 102.86V
r =
0.175
The potential energy is ∆P E = q∆V = ke q1rq2 = −3.85&middot;10−7 , where the algebraic sign, which is negative,
means that the two charges want to come closer to each other (we must intervene and push them apart to
keep them separated).
Problem 25
The electric field is given by E =
V
20
d = 1.8&middot;10( −3)
0 A
d with 0 =
= 1.1 &middot; 104 N/C. Field lines point from positive to negative
(always). The capacitace is C =
8.85 &middot; 10−12 , which gives C = 3.74pF . Lastly the charge is
given by Q = CV , which gives Q = 74.7pC on each plate.
Problem 31
First, add the two capacitors in parallel, which add directly to a 6&micro;F capacitor. Then, add that one in series
with the remaining capacitor. The equivalent capacitance of the circuit is 2&micro;F . The total charge in the
circuit is then Qtot = Ceq V = 24&micro;C. Going back one step in the reduction process (one 3&micro;F and one 6&micro;F
in series), both of these have the same charge (24&micro;C). Capacitors in series always have the same charge on
them. Thus we have solved the 3&micro;F capacitor:It has a charge of 24&micro;C and a voltage of V = Q
C = 8V . Going
back one more step, the 6&micro;F capacitor also has 24&micro;C on it, which means it has a voltage of V = Q
C = 4V ,
but remember this is a composite capacitor (it’s really made up of two capacitors). Since the two constituent
1
capacitors are in parallel, both of these capacitors have the same voltage, namely 4V. The charge on each
one is then given by Q = CV . The 4&micro;F capacitor has 16&micro;C and the 2&micro;F capacitor has 8&micro;C.
2
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