Part (b) Find the electric potential, in kilovolts, at the origin due solely to the charge on the bottom side of the loop by integrating the
infinitesimal contributions to the potential from the side’s infinitesimal segments.
We consider an infinitesimal segment of the bottom side of the loop. The segment is located at (x, − L2 ) and has length dx. We use the
expression for the linear charge density that we found in part (a) to write the infinitesimal charge on this segment,
dQ =
Qdx
4L.
Pythagoras tells us that the distance of this infinitesimal segment from the origin is
−−−−−−−−−2−
L
) .
2
d = √x2 + (
The contribution of the infinitesimal charge on this segment to the potential at the origin is
dV =
1 dQ
1
=
4πϵ0 d
4πϵ0
Qdx
−−−−−−−−2 .
4L√x2 + ( L2 )
To find the potential at the origin due to the whole bottom side of the loop, we integrate the expression for dV from x = − L2 to x = + L2
(an integral table might be helpful here) and obtain
V1 =
1 Q
ln (1 + √–
2) .
4πϵ0 2L
We substitute the given values, taking into account the given length unit and the requested unit of potential,
V1 =
–
25 × 10−6 C ln (1 + √2)
1
.
4πϵ0 2 (5.5 ⋅ 0.01 m)
1000
V1 = 1800. kV
Part (c) Use symmetry to determine the electric potential at the origin due to the entire loop. Give your answer in kilovolts.
The charge distribution possesses 4-fold rotation symmetry about the z-axis. Since potential is a scalar quantity (doesn't have a direction),
each side of the square loop thus contributes equally to the potential at the origin. Therefore, the potential at the origin due to the entire
loop is four times the result we found in part (b),
V = 4V1 = 4 (1800. kV) .
V = 7202 kV
Problem 23:
An initially neutral conducting sphere has a radius R = 0.2 m. Electrons are fired towards the sphere with an initial
speed of v = 2 × 106 m/s. Electrons striking the surface remain on the sphere and a net negative charge accumulates
over time. Assume the electrons are fired far from the sphere.
Part (a) How many electrons will be deposited on the sphere before they no longer reach the surface?
The electrons are shot from a very far distance (infinity) at the sphere. The first one doesn't lose any speed until it impacts the sphere. The
second one will see the sphere having the charge of one electron and will lose some energy before impacting the sphere. Eventually, the
energy loss will be just enough that the electron stops before impacting the sphere. This will occur when the initial kinetic energy of the
electron is equal to the potential energy of the electron-sphere configuration at a separation of R. The kinetic energy of the electron is
Ek =
1
me v2 J
2
where me is the mass of the electron in kg and v is the speed in m/s. Since the conductor is a sphere, it can be treated as a point charge
with all its charge located at the center. The potential energy between two point charges is
Ep =
qQ
J
(4πϵ0 r)
where q is the charge of one of the point charges in C, Q is the charge of the other point charge in C, r is the separation in m, and ϵ0 is the
vacuum permittivity in C2 /m2 /N. Choosing q to be the charge of the fired electron and Q as the charge of the sphere,
Ep =
q (nq)
(4πϵ0 r)
Equating the energies and solving for the number of electrons, n,
q (nq)
1
me v2 =
2
(4πϵ0 r)
q 2 n = (4πϵ0 r) ⋅
1
me v2
2
v 2
n = 2πϵ0 rme ( )
q
Plugging in numbers
n = 2 ⋅ π ⋅ 8.854188 ⋅ 10−12 C2 m−2 N−1 ⋅ 0.2 m ⋅ 9.109384 ⋅ 10−31 kg ⋅ (
(2 ⋅ 106 m/s)
(1.602177 ⋅ 10−19 C)
)
2
n = 1.580 × 109 electrons
Problem 24:
A parallel-plate capacitor is constructed from two square sheets of aluminum foil, each of dimensions 0.15 m × 0.15
m. The plates are separated by an air gap of 1.5 mm.
Part (a) Calculate the capacitance of the capacitor, in picofarads.
The capacitance of an air filled parallel-plate capacitor is given by
C = ϵ0
A
,
d
where A denotes the area of each plate and d is the plate separation. Since the problem gives us the side length L of each square plate, we
rewrite the capacitance formula as
C = ϵ0
L2
.
d
We substitute the given values, taking into account the given length units and the requested unit for the capacitance,
C = ϵ0
(0.15 m)2
1012 .
1.5 ⋅ 0.001 m
C = 132.8 pF
Part (b) What will the capacitance be, in picofarads, if the plate separation is doubled?
Since the capacitance is inversely proportional to the plate separation, doubling the plate separation will reduce the capacitance to half its
original value. We use our result from part (a) to obtain
C=
1
(132.8 pF) .
2
C = 66.40 pF
Part (c) Calculate what the capacitance will be, in picofarads, if the plates are halved in area while keeping the original plate
separation.
Since the capacitance is proportional to the plate area, halving the plate area while keeping the original plate separation will again reduce
the capacitance to half its original value. We again use our result from part (a).
C=
1
(132.8 pF) .
2
C = 66.40 pF
Problem 25:
Three capacitors are connected in series as shown in the figure. The capacitances are C1
= 2.1 μF, and C2 = 4.2 μF, C3 is unknown, and the charge stored in each capacitor is Q
= 2.5 μC.
Part (a) Express the capacitance C of a capacitor in terms of charge Q and voltage ΔV on it.
We learn from our textbook that the capacitance
C
of a capacitor, in terms of its charge
Q
and the voltage
ΔV
on it, is
Q
C = ΔV
Part (b) Apply the above formula to capacitor C1 to find an expression for the potential difference ΔV12 across it.
We use the expression we showed in part (a),
Q
C = ΔV ,
substitute the relevant symbols,
Q
C1 = ΔV
12
,
and solve for
ΔV12
,
Q
V12 = C
1
Part (c) Express the potential difference ΔV12 through potentials V1 and V2 where V1 and V2 are the potentials measured in the wires 1
and 2, respectively, relative to the negative side of the battery.
The requested potential difference
ΔV12
is the difference between the potentials
V1
and
V2
. We are told that the reference for potentials is the battery's negative terminal, so the potentials indicated in the diagram are all positive and
increase from the negative terminal to the positive one. Thus, we write
ΔV12
as
V12 = V1 − V2
Part (d) Calculate V2 in V given V1 = 9 V.
We use the expression we found in part (c),
V12 = V1 − V2 ,
and solve it for
V2
,
V2 = V1 − V12 .
We substitute the expression for
ΔV12
that we found in part (b),
Q
V12 = C ,
1
and obtain
Q
V2 = V1 − C .
1
We substitute the given values, taking into account the given units for charge and capacitance,
−6
V2 = (9 V) − 2.5×10−6 C .
2.1×10
F
V2 = 7.810 V
Part (e) Repeat the above procedure for capacitor C2 and calculate the potential at point 3, V3 in V.
We apply the same method that we used to find
V2
in part (d) and write
Q
V3 = V2 − C .
2
We substitute the expression for
V2
that we found in part (d),
Q
V2 = V1 − C ,
1
and obtain
V3 = V1 − Q ( C1 + C1 ) .
1
2
We substitute the given values, taking into account the given units for charge and capacitance,
V3 = (9 V) − (2.5 × 10−6 C) (
V3 = 7.214 V
1
1
+
) .
2.1×10 −6 F
4.2×10 −6 F
Problem 26:
Suppose you want a capacitor bank with a total capacitance of 0.615 F and you possess numerous 1.5 mF capacitors.
Part (a) What is the smallest number you could connect together to achieve your goal?
In order to achieve a higher capacitance with a number of identical lower-capacitance capacitors, the latter must be combined in parallel.
When we combine
N
capacitors, each of capacitance
C
, in parallel, the equivalent capacitance
Ceq
of the combination will be
Ceq = NC .
We solve for
N
,
C
N = Ceq ,
and substitute the given values, taking into account the given units,
N=
0.615 F
.
1.5×10 −3 F
N = 410.0
Problem 27:
The diagram shows three capacitors, an ideal battery, and an open switch, S. The three
capacitors all have the same capacitance. Determine what happens to the following
quantities after the switch has been closed.
Part (a) Equivalent capacitance of a circuit after the closing of switch S:
Before the switch is closed, we have two capacitors, both with capacitance C , in series. This means that the capacitance of the circuit is
given by
1
1
1
=
+
Ct1
C
C
1
2
=
Ct1
C
Ct1 =
1
C
2
Now, if we close the switch, we add C3 to the system. Let's first find out the equivalent capacitance of the parallel connection between C2
and C3 .
C23 = C + C
C23 = 2C
Now we can put the combined value we just found in series with C1 and find an overall equivalent capacitance for the circuit.
1
1
1
=
+
C
Ct2
C23
1
1
1
=
+
C
2C
Ct2
1
3
=
Ct2
2C
Ct2 =
2
C
3
A quick comparison tells us that the capacitance of the circuit with the closed switch is higher. As such, the answer is that the capacitance
Increases
Part (b) Total charge in this circuit after closing the switch S:
In part (a), we found values for the capacitance of the circuit with the switch open (Ct1 ) and closed (Ct2 ). We also know that the voltage
across each of the combined capacitors must be the same, as they are both being driven by the same battery. Now, let's use the relation
between voltage, capacitance, and charge to find a value for the charge of Ct1 .
Ct1 =
q
1
C = t1
V
2
1
CV = qt1
2
Now, let's do the same for Ct2 .
Ct2 =
qt2
2
C=
V
3
2
CV = qt2
3
Comparing the two results, we can see that the charge on the equivalent capacitor of the closed-switch system is higher. As such, the
answer is that the charge on the circuit
Increases
Part (c) Energy stored in this circuit after closing the switch S:
In part (b), we found values for the charge of the circuit with the switch open (qt1 ) and closed (qt2 ). We also know that the voltage across
each of the combined capacitors must be the same, as they are both being driven by the same battery. Now, let's use the equation for the
energy of a capacitor to find the energy stored in the system with the open switch.
Ut1 =
1
qt1 V
2
Ut1 =
1
CV 2
4
Now, let's do the same for the circuit with the closed switch.
Ut2 =
1
qt2 V
2
Ut2 =
2
CV 2
6
Ut2 =
1
CV 2
3
Comparing the two results, we can see that the energy stored by the equivalent capacitor of the closed-switch system is higher. As such,
the answer is that the energy stored by the circuit
Increases
Part (d) Charge on capacitor 1 after closing the switch S:
Before closing the switch, we had capacitors C1 and C2 in series by themselves. Remember that for capacitors in series, they must each
have the same amount of charge as each other and the equivalent capacitor. We can therefore refer to our results from part (b) and note that
the charge on C1 with the switch open, denoted by qC1t1 , is given by:
qC1t1 = qt1
qC1t1 =
1
CV
2
Now, we can apply the same logic when the circuit is closed. That is to say, since C1 and the equivalent capacitor of C2 and C3 , C23 , are
in a series together, they must each have the same charge as the equivalent capacitor. Again referring to our results from part (b), we can
note the charge on C1 with the switch closed, qC1t2 , is equal to:
qC1t2 = qt2
qC1t2 =
2
CV
3
From this, we can clearly see that the charge on the first capacitor is higher when the circuit is closed. As such, the charge on the capacitor
Increases
Part (e) Voltage drop on capacitor 1 after closing the switch S:
Since the capacitance of a capacitor is determined by its geometry, we know that the capacitance of C1 is still the same. Since we found
the charges on C1 when the switch is open (qC1t1 ) and closed (qC1t2 ) and we know that the capacitance, we can find the voltage. Let's start
with the voltage when the switch is open, denoted by VC1t1
VC1t1 =
qC1t1
C
VC1t1 =
1 CV
2 C
VC1t1 =
1
V
2
Now let's do the same to find a voltage for when the switch is closed, letting the voltage across C1 in this case be denoted by VC1t2 .
VC1t2 =
qC1t2
C
VC1t2 =
2 CV
3 C
VC1t2 =
2
V
3
From this, we can see that the voltage drop across C1 is clearly larger when the switch is closed. As such, the answer is that it
Increases
Part (f) Energy stored in capacitor 1 after closing the switch S:
In parts (d) and (e), we found that both the charge and the voltage increased when the switch was closed. Now, let's look at the equation
for the energy stored on a capacitor.
U=
1
qV
2
Since the energy is dependent upon both the charge and voltage, if both are higher with the switch closed, then energy must logically be
higher with the switch closed as well. As such, we know that the energy stored on this capacitor
Increases
Part (g) Charge on capacitor 2 after closing the switch S:
We found values for the amount of charge on C1 with the switch open and with the switch closed in part (d). Since capacitors in parallel
have the same amount of charge on them, we can state that C2 's charge is the same as C1 's charge with the switch open. Letting the charge
on C2 with the switch open be denoted qC2t1 , we can state
qC2t1 =
1
CV
2
Now let's examine the scenario where the switch is closed. Since the combined capacitor made of C2 and C3 , C23 , is in series with C1 ,
they must have the same charge. Thus, we know the charge q23 from part (d) as well.
q23 =
2
CV
3
Now, since C2 and C3 are in parallel with one another, they must have the same voltage. Further, we know from the problem statement
that they have the same capacitance. As a consequence, their charges must be equal as well. Since charge is conserved, if they have equal
charges, that means that C2 and C3 must each get half the total available charge on C23 . Using this information, we can find the total
charge on C2 with the switch open, which we will denote as qC2t2 .
qC2t2 =
1
q23
2
qC2t2 =
1 2
( CV )
2 3
qC2t2 =
1
CV
3
We thus see that the overall charge on C2 is lower with the switch closed than it was when the switch was open. Therefore, the answer is
that the charge
Decreases
Part (h) Voltage drop on capacitor 2 after closing the switch
In part (g), we found the charge on C2 when the switch is both open and closed. Using the relation between charge, voltage, and
capacitance of a capacitor, let's first find the voltage drop with the switch open which we will denote VC2t1 .
VC2t1 =
qC2t1
C
VC2t1 =
1 CV
2 C
VC2t1 =
1
V
2
Now let's repeat this process to find the voltage drop when the switch is closed, denoted by VC2t2 .
VC2t2 =
qC2t2
C
VC2t2 =
1 CV
3 C
VC2t2 =
1
V
3
Since the voltage drop clearly has a lower value with the switch closed, we can say that it
Decreases
Part (i) Energy stored in capacitor 2 after closing the switch:
In parts (g) and (h), we found that both the charge and the voltage on C2 decreased when the switch was closed. Now, let's look at the
equation for the energy stored on a capacitor.
U=
1
qV
2
Since the energy is dependent upon both the charge and voltage, if both are lower with the switch closed, then energy must logically be
lower with the switch closed as well. As such, we know that the energy stored on this capacitor
Decreases
Problem 28:
A cylindrical capacitor is made of two concentric conducting cylinders. The inner cylinder has radius R1 = 19 cm
and carries a uniform charge per unit length of λ = 30 μC/m. The outer cylinder has radius R2 = 25 cm and carries an
equal but opposite charge distribution as the inner cylinder.
Part (a) Use Gauss’ Law to write an equation for the electric field at a distance R1 < r < R2 from the center of the cylinders. Write your
answer in terms of λ, r, and e0.
Let's consider a Gaussian surface that is a cylinder surrounding the inner cylinder of the capacitor. No flux will be passing through the flat
ends of our cylindrical Gaussian surface, so we only need to consider the curved surface. Considering the surface area of the relevant part
of our Gaussian surface and the charge on the inner cylinder of the capacitor, we can write the following equation using the symmetry of
the system:
EA =
Q
ϵ0
E (2πrl) =
E=
λl
ϵ0
λ
2πϵ0 r
Part (b) Calculate the electric potential difference between the outside and the inside cylinders in V.
To find the change in voltage between the two plates, we will need to integrate the electric field along a path between the two plates. Using
our answer from part (a), we can perform the desired integration as follows:
ΔV = ∫ E ⋅ d s
ΔV = ∫
R2
R1
(
λ
)⋅dr
2πϵ0 r
ΔV =
λ
⋅ (−ln (R 1 ) + ln (R 2 ))
2πϵ0
ΔV =
λ
R2
⋅ ln (
)
2πϵ0
R1
Now we can plug in values and solve. As we do so, let's be careful to convert the charge per unit length to coulombs per meter. Since we
are dividing the two lengths by one another, the units will cancel meaning that we do not need to convert them from centimeters to meters
in this case.
ΔV =
30 ⋅ 10−6 C/m
2π ⋅ 8.85418 ⋅ 10−12 F/m
⋅ ln (
25 cm
)
19 cm
ΔV = 1.480 × 105 V
Part (c) Calculate the capacitance per unit length of these concentric cylinders in F/m.
To begin, let's write the relationship between capacitance, charge, and voltage.
C=
q
V
In this case, we want to find the capacitance per unit length, so we need to divide both sides by the length of the cylinder to get the correct
equation. Further, the total charge on the capacitor is equal to its length multiplied by the charge per unit length. This gives us the
following equation:
C/ l =
λl
Vl
C/ l =
λ
V
To proceed from here, we simply need to use the equation we found in part (b) for the potential difference in the capacitor and plug in
values to solve.
C/ l =
C/ l =
λ
R
λ
( 2πϵ
⋅ ln ( R2 ))
0
1
2πϵ0
ln ( R2 )
R
1
C/ l =
2π ⋅ 8.85418 ⋅ 10−12 F/m
cm
ln ( 25
)
19 cm
C/ l = 2.027 × 10−10 F/m
Part (d) Calculate the energy stored in the capacitor per unit length, in units of J/m.
Let's begin by writing out the equation relating the stored energy of a capacitor to its capacitance and voltage.
U=
1
CV 2
2
As we are being asked to find the energy stored per unit length, let's divide both sides by the length.
U/ l =
1C 2
V
2 l
In parts (b) and (c), we found expressions for the voltage across the capacitor and the capacitance per unit length. We can now plug these
expressions in to our current equation and solve. Note that, while it isn't necessary to convert the radii from centimeters to meters as their
units will cancel out in division, we do need to convert the charge per unit length from microcoulombs per meter to coulombs per meter.
U/ l =
2
1 ⎛ 2πϵ0 ⎞
λ
R2
⋅ ln (
))
⎜
⎟(
2 ⎝ ln ( R2 ) ⎠ 2πϵ0
R1
R1
U/ l =
1 λ2
R2
⋅ ln (
)
2 2πϵ0
R1
U/ l =
λ2
R2
⋅ ln (
)
4πϵ0
R1
2
U/ l =
(30 ⋅ 10−6 C/m )
−12 F/m
4π ⋅ 8.85418 ⋅ 10
⋅ ln (
25 cm
)
19 cm
U/ l = 2.220 J/m
Part (e) Write an equation for the energy density due to the electric field between the cylinders in terms of λ, r, and e0.
Let's begin by writing the equation for the energy density of an electric field.
u=
1
ϵ0 E 2
2
We can now plug in the equation we found for the electric field in part (a) and simply to find the answer.
u=
u=
2
1
λ
)
ϵ0 (
2πϵ0 r
2
λ2
8π 2 ϵ0 r2
Part (f) Consider a thin cylindrical shell of thickness dr and radius R1 < r < R2 that is concentric with the cylindrical capacitor. Write
an equation for the total energy per unit length contained in the shell in terms of λ, r, dr, and ε0.
The energy contained in the shell will be equal to the energy density multiplied by the volume of the shell within the electric field. As
noted in part (a), there is no electric flux (and thus, no electric field) passing through the flat sides of the cylinder. Thus, we need to
multiply the volume of the curved section of our cylindrical shell by the energy density to find the energy contained in the shell.
d U = Au
Now, let's write out the volume of the curved portion of a cylinder and plug in the expression we found for the energy density in part (e).
d U = (2πrld r) (
dU =
λ2
)
8π 2 ϵ0 r2
λ2 ld r
4πϵ0 r
Since we were asked for the energy per unit length, we can now divide both sides by the length in order to get the desired equation.
d U/ l =
λ2 d r
4πϵ0 r
Part (g) Calculate the energy stored per unit length in the capacitor in units of joules per meter.
To find the total energy stored in the electric field, we can integrate the equation we found in part (f) over the length of the electric field.
Let's begin by considering the limits of our integration. Since the inner cylinder is a conductor, there is no electric field inside it. Further,
since the charge on the outer cylinder is equal and opposite that of the inner cylinder, the net charge contained in a Gaussian field
surrounding the entire capacitor would be zero. As such, the only region with an electric field is the space between the two capacitors.
With this in mind, let's integrate the equation we found in part (f) from the surface of the inner cylinder out to the outer cylinder.
λ2 d r
4πϵ0 r
d U/ l =
U/ l = ∫
R2
R1
λ2 d r
4πϵ0 r
U/ l =
λ2
(−ln (R 1 ) + ln (R 2 ))
4πϵ0
U/ l =
λ2
R
ln ( 2 )
4πϵ0
R1
We can now plug in values and solve, making sure to convert the charge per unit length from microcoulombs per meter to coulombs per
meter. As in previous parts of this problem, we do not need to convert the radii of the cylinders as their units will cancel out when they
divide each other.
2
(30 ⋅ 10−6 C/m )
U/ l =
−12
4π ⋅ 8.85418 ⋅ 10
F/m
⋅ ln (
25 cm
)
19 cm
U/ l = 2.220 J/m
Problem 29:
Suppose you have a 9.00 V battery, a 2.2 μF capacitor, and a 7.25 μF capacitor.
Part (a) Find the total charge stored in the system if the capacitors are connected to the battery in series in C.
The equivalent capacitance
C
of a series combination of two capacitors with capacitances
C1
and
C2
is given by
1
C
= C1 + C1 ,
1
2
C C
C = C 1+C2 .
1
2
The requested charge
Q
stored in the capacitors when they are connected to a battery producing voltage
V
is
C C V
Q = CV = C1 +2C .
1
2
We substitute the given values, taking into account the given capacitance unit,
Q=
(2.2×10 −6 F)(7.25×10 −6 F)(9.00 V)
(2.2×10 −6 F)+(7.25×10 −6 F)
.
Q = 1.519 × 10−5 C
Part (b) Find the energy stored in the system if the capacitors are connected to the battery in series in J.
The energy
Us
stored in the series-connected capacitors, in terms of their equivalent capacitance
C
, their individual capacitances
C1
and
C2
, and the battery voltage
V
, is
C C V2
Us = 12 C V 2 = 12 C1 +2C .
1
2
We substitute the given values, taking into account the given capacitance unit,
Us = 12
2
(2.2×10 −6 F)(7.25×10 −6 F)(9.00 V)
(2.2×10 −6 F)+(7.25×10 −6 F)
.
Us = 6.836 × 10−5 J
Part (c) Find the charge if the capacitors are connected to the battery in parallel in C.
The equivalent capacitance
C
of a parallel combination of two capacitors with capacitances
C1
and
C2
is given by
C = C1 + C2 .
The requested charge
Q
stored in the capacitors when they are connected to a battery producing voltage
V
is
Q = CV = (C1 + C2 ) V .
We substitute the given values, taking into account the given capacitance unit,
Q = ((2.2 × 10−6 F) + (7.25 × 10−6 F)) (9.00 V) .
Q = 8.505 × 10−5 C
Part (d) Find the energy stored if the capacitors are connected to the battery in parallel in J.
The energy
Up
stored in the parallelly connected capacitors, in terms of their equivalent capacitance
C
, their individual capacitances
C1
and
C2
, and the battery voltage
V
, is
Up = 12 C V 2 = 12 (C1 + C2 ) V 2 .
We substitute the given values, taking into account the given capacitance unit,
Up = 12 ((2.2 × 10−6 F) + (7.25 × 10−6 F)) (9.00 V)2 .
Up = 3.827 × 10−4 J
Problem 30:
Consider a parallel plate capacitor of capacitance C and distance between plates d and plate area 2A. A slab
of dielectric material of thickness d and dielectric constant κ>1 is inserted between plates filling half of
region between the plates (see figure).
Part (a) The capacitance of such “combined” capacitor is now:
For our initial capacitor, we can find a value for the capacitance based on the geometry. Given that no dielectric is present initially, we get
the following equation for the capacitor before a dielectric is added.
C=
ϵ0 (2A)
d
C=
2ϵ 0 A
d
Now, let's examine the system once we have added the dielectric. Once the dielectric is added, we can view the capacitor as two capacitors
in parallel. One of these parallel capacitors would have a dielectric and the other would not. We know that the distance between the plates
remains the same and that each of these new capacitors would have half the area of the original capacitor. Using the equation for the
capacitance of capacitors in parallel, let's write an equation for the capacitance after the dielectric is added and work through it to find an
answer.
Cf =
κϵ A
ϵ0 A
+ 0
d
d
Cf =
ϵ 0 A + κϵ 0 A
d
Cf =
ϵ0 A (κ + 1)
d
Cf =
ϵ0 A
(κ + 1)
d
Cf =
1
C (κ + 1)
2
Cf =
(κ + 1) C
2
This gives us a final answer of
(κ + 1) C
2
Problem 31:
Full solution not currently available at this time.
The figure shows the equipotential
surfaces in some region of space. The
question below refer to the points marked
A-F in the figure.
Part (a) At which of the following points is the electric field the strongest?
E
Part (b) At which of the following points would an electron have the greatest potential energy?
E
Part (c) What is the potential difference between points C and A?
ΔVC→A = -70
ΔVC→A = −70.00
Tolerance: ± 2.1
Part (d) Which best describes the direction of the electric field at point B?
directly to the left
Part (e) How much work would be required to move a charge of q = -4 μC from point A to F to E?
80.00 μJ
Problem 32:
Full solution not currently available at this time.
The graph in the figure shows the variation
of potential, V, along the x-axis.
Part (a) The slope of the tangent at point B represents
The magnitude of the electric field at x = 2 m
Part (b) At point A, the electric field...
...points to the left
Part (c) At point C, the electric field...
...points to the left
Part (d) At point D, the electric field...
...points to the right
Part (e) Which of the following best represents the electric field along the x-axis?
Problem 33:
A thin rod of length L=50 cm carries a total charge of Q = -3.2 nC uniformly distributed over its right half and +3.2
nC on the left half. Find the electric potential on the axis of the rod at distance r=60 cm from its rightmost end.
Part (a) Select all that are true about the potential at point P:
negative
Part (b) Write an expression for the integrand, in terms of dx, Q, k, r, and L, that is dV, the electric potential at point P.
Part (c) What is the electric potential at point P. (in V)? In your work, be sure to write the final expression for V before plugging in
numbers and also use GOAL protocol.
Problem 34:
A parallel place capacitor is made of square plates that are
length, L on each side is used to measure the unknown
dielectric constant of a material, κ2.
The plate separation, d, is filled with two dielectric materials
as shown in the figure. The material on the left has a known
dielectric constant of κ1.
Part (a) When the plates are connected to a voltage source, V as shown in the figure, the charge on the plates is measured to be, Q.
Express the unknown dielectric constant,κ2, in terms of the values given in the problem
The obvious question that arises here is "how do we treat the different dielectrics?".
If you look at the connection to the circuit, you can see that the entire top plate is at the potential of the positive battery terminal, and the
entire bottom plate is at the potential of the negative battery terminal.
Since both halves of the capacitor are at the same voltage, this is equivalent to two capacitors connected in parallel (see figure):
2
We can treat this capacitor as two capacitors with sides L and L2 (and therefore plate area A = L2 ), and plate separation d that are
connected in parallel to the same battery, V .
L2
⇒ C1 =
κ1 ϵ0 2
d
L2
and C2 =
κ2 ϵ0 2
d
⇒ Ceq = C1 + C2 =
ϵ0 L2
(κ1 + κ2 )
2d
Q
Given that the charge of the entire system is Q , we can set this equal to V and solve for κ2 :
Ceq =
ϵ0 L2
Q
(κ1 + κ2 ) =
2d
V
⇒ (κ1 + κ2 ) =
⇒
κ2 =
2dQ
ϵ0 L2 V
2dQ
− κ1
ϵ0 L2 V
Problem 35:
Full solution not currently available at this time.
An air filled, parallel plate capacitor is connected to a battery supplying constant voltage, V. The energy stored in the
capacitor is calculated to be U0.
What is the new energy stored in the capacitor after each of these changes? (Each change starts from the initial
conditions, they are not sequential).
Part (a) A dielectric of constant, κ is inserted between the plates of the capacitor. Assume that it completely fills the space between the
plates.
κU0
Part (b) The separation of the plates of the capacitor is halved
2U0
Part (c) The battery is disconnected from the capacitor.
U0
Part (d) The battery is disconnected from the capacitor and then a dielectric of constant, κ is inserted between the plates of the
capacitor. Assume that it completely fills the space between the plates.
U0
Part (e) The battery is disconnected from the capacitor and then the separation of the plates of the capacitor is halved
U0/2
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