Questions:  1  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20 
(a) Since the plate is a conductor we know that all charge must reside on it's two sides. Let's assume that the charge one side is Q_{1} and Q_{2} on the other. The charge densities are then Σ_{1} = Q_{1}/A and Σ_{2} = Q_{2}/A. The electric field due to the rightside plane of charge is E_{1} = Σ_{1}/2ε_{0} and due to the leftside plane of charge is E_{2} = Σ_{2}/2ε_{0}, as shown in the diagram below. The field due to the left side is shown in solid red while the field due to the to the right side is dashed green. Both sides are assumed to have positive charge.
According to the diagram the net field inside the conductor is
E_{net} = E_{1}  E_{2} .
We know that the E_{net} = 0 since the slab is a conductor. Thus, using our equations for the field due to a plane of charge
0 = Σ_{1}/2ε_{0}  Σ_{2}/2ε_{0} .
This yields the result that Σ_{1} = Σ_{2}, so the sides have equal charge densities. Since Q = ΣA, and the sides have the same are, then Q_{1} = Q_{2} and the sides have the same charge. From Conservation of Charge, we know that Q_{1} + Q_{2} = Q, hence Q_{1} = Q_{2} = ½Q. Thus the charge is spread equally on the two surfaces.
(b) When the plates are brought close together, the charge distribution on one plate will change from an equal distribution on each side as in part (a) because of the electric field of the other plate. Let Σ_{L} = ½(Q_{1}/A) be the initial charge density on either side of the left plate and Σ_{R} = ½(Q_{2}/A) on the sides of the right plate. If we assume that charge density on the right side of the left plate changes by an amount Ds_{L}, the charge density of the left side must decrease by a similar amount by conservation of charge. Similarly the charge on each side of the right plate increases or decreases by Ds_{R}. This is shown in the sketch below.
Now, the field must be horizontal by symmetry as long as we are not near the edges of the plates. The field due to each side is given by E = Σ/2ε_{0}. We assume that the field for the left plate are positive and those from the right plate are negative, we get E_{1} = (Σ_{L}Ds_{L})/2ε_{0}, E_{2} = (Σ_{L}+Ds_{L})/2ε_{0}, E_{3} = (Σ_{R}+Ds_{R})/2ε_{0}, and E_{4} = (Σ_{R}Ds_{R})/2ε_{0}, as shown in the sketch below.
The net electric fields in the various regions are
E_{A}  =  E_{1} + E_{2} + E_{3} + E_{4}  =  (2Σ_{L}  2Σ_{R})/2ε_{0}  (1) 
E_{B}  =  E_{1} + E_{2} + E_{3} + E_{4}  =  (2Ds_{L}  2Σ_{R})/2ε_{0}  (2) 
E_{C}  =  E_{1} + E_{2} + E_{3} + E_{4}  =  (2Σ_{L}  2Σ_{R})/2ε_{0}  (3) 
E_{D}  =  E_{1} + E_{2}  E_{3} + E_{4}  =  (2Σ_{L} + 2Ds_{R})/2ε_{0}  (4) 
E_{F}  =  E_{1} + E_{2}  E_{3}  E_{4}  =  (2Σ_{L} + 2Σ_{R})/2ε_{0}  (5) 
The field inside each plate, E_{B} and E_{D}, must be zero. Therefore equation (2) and equation (4) become
2Ds_{L}  2Σ_{R} = 0 ,
and
2Σ_{L} + 2Ds_{R} = 0 .
Thus we have Dss_{L} = Σ_{R} and Ds_{R} = Σ_{L}. Thus the total charge density on each plate is Σ_{L}+Σ_{R}, Σ_{L}Σ_{R}, Σ_{R}Σ_{L}, and Σ_{R}+Σ_{L} as we go from left to right. Thus the inner faces have equal but opposite charge densities while the outer sides have equal charge densities. The charge on each plate is Q = ΣA, so the charge on each side of each plate is (Σ_{L}+Σ_{R})A = ½(Q_{1} + Q_{2}), (Σ_{L}Σ_{R})A = ½(Q_{1}  Q_{2}), (Σ_{R}Σ_{L})A = ½(Q_{2}  Q_{1}), and (Σ_{R}+Σ_{L})A = ½(Q_{1} + Q_{2}) as we go from left to right. Thus the inner faces have equal but opposite charges while the outer sides have equal charges.
From equations (1), (3), and (5), the electric fields would then be
E_{left} = (Σ_{L} + Σ_{R})/ε_{0} = (Q_{1} + Q_{2})/2Aε_{0} ,
E_{centre} = (2Σ_{L}  2Σ_{R})/2ε_{0} = (Q_{1}  Q_{2})/2Aε_{0} ,
and
E_{right} = +(Σ_{L} + Σ_{R})/ε_{0} = (Q_{1} + Q_{2})/2Aε_{0} .
The left and right electric fields are equal but opposite.
(a) Using the results from question 1, the charge on the sides as we go from left to right are ½[Q + (Q)] = 0, ½[Q_{1}  (Q_{2})] = +Q, ½[(Q)  Q)] = Q, and ½[Q + (Q)] = 0. So in this case all the charge is on the inner faces and there is no charge on the outer faces. Since a simple capacitor is just two thin plates, we can assume that all the charge on a capacitor resides on the inner sides as well.
(a) Using the results from question 1, the electric fields would be
E_{left}  =  [Q + (Q)]/2Aε_{0} = 0 
E_{centre}  =  [Q  (Q)]/2Aε_{0} = Q/Aε_{0} 
E_{right}  =  [Q + (Q)]/2Aε_{0} = 0 
(a) Using the results from question 1, the charge on the sides as we go from left to right are ½[(7.0 μC)+(5.0 μC)] = +1.0 μC, ½[(7.0 μC)(5.0 μC)] = +6.0 μC, ½[(5.0 μC)  (7.0 μC)] = 6.0 μC, and ½[(7.0 μC) + (5.0 μC)] = +1.0 μC. So in this case all the charge is on the inner faces and an equal charge on the outer faces.
(b) Using the results from question 1, the electric fields would be
E_{left}  = [(7.0 μC) + (5.0 μC)]/2(5.0 m^{2})ε_{0} = 2.26 × 10^{4} V/m 
E_{centre}  = [(7.0 μC)  (5.0 μC)]/2(5.0 m^{2})ε_{0} = 1.36 × 10^{5} V/m 
E_{right}  = [(7.0 μC) + (5.0 μC)]/2(5.0 m^{2})ε_{0} = 2.26 × 10^{4} V/m 
According to the results of question 2, the charge resides on the inner faces of the capacitors as shown below.
Before S is closed all the charge is on the inside plates of each capacitor. Since closing S connects two outer uncharged plates, there is no potential difference and one would not expect any charge to flow. However let us assume that charge has moved and that the charge on the two connected plates has changed. The results of question 1 indicate that when two plates have a different amount of charge, the arrangement of charge must be such that the charges on the inner plates of the capacitor are equal but opposite while the outer faces must have the same charge. This is shown in the diagram below.
Conservation of charge requires that
Q_{1} + Q_{2}  = Q_{A}  (1) 
Q_{2} + Q_{1} + Q_{3} + Q_{4}  = Q_{A} + Q_{B}  (2) 
Q_{4} + Q_{3}  = Q_{B}  (3) 
Furthermore, for the system to be in equilibrium with no charge flowing,
Q_{1}  = Q_{3}  (4) 
Using equation (4) to eliminate Q_{3} from the first three equations we get
Q_{1} + Q_{2}  = Q_{A}  (1) 
Q_{2} + 2Q_{1} + Q_{4}  = Q_{A} + Q_{B}  (2) 
Q_{4} + Q_{1}  = Q_{B}  (3) 
Adding equations (1), (2), and (3) together, we get
4Q_{1}  = 0  (4) 
So the charges on the outer plates is still zero, Q_{2} = Q_{A}, and Q_{4} = Q_{B}. Thus there has been no redistribution of charge as we had predicted.
The results of question 1 show that the only arrangement of charge which keeps the electric field equal to zero inside the plates is with the inner faces have equal but opposite charge and the outer faces having the same charge, as shown below.
Let's assume that the lower inner plates had charge +Q_{A} and +Q_{B} respectively and then the upper plates would have had charge Q_{A} and Q_{B}. Conservation of charge says that
Q_{2} + Q_{1} + Q_{3}  Q_{4}  = Q_{A}  Q_{B}  (1) 
Q_{2} + Q_{1} + Q_{3} + Q_{4}  = Q_{A} + Q_{B}  (2) 
If we add equations (1) and (2) together, we get
2Q_{1} + 2Q_{3}  = 0  (3) 
Since the capacitors must be in equilibrium, Q_{1} must equal Q_{3} if no charge is to flow. Thus equation (3) implies that
Q_{1} = Q_{3} = 0 .
Thus all the charge is on the inner faces and equations (2) and (3) reduce to
Q_{2} + Q_{4} = Q_{A} + Q_{B} .
As well, it is necessary that Q_{2} = Q_{A} and Q_{4} = Q_{B} if the voltage drop over each capacitor is to remain unchanged.
Capacitance is defined as C = Q/ΔV, where ΔV = ò_{+}^{} E•dl .
Since we know the electric field, we can determine ΔV and thus C. We choose the line integral to run along the x direction since the electric field points in this direction. Our integral then reduces to
DV = ò_{+}^{} E•dl = ò_{0}^{d} Edx = Σd/ε_{0} ,
where we take the positive plate as x = 0.
Note that Q = ΣA, so
C = ΣA / Σd/ε_{0 }= ε_{0}A /d .
As expected, the capacitance depends only on the geometrical factors A and d and on a constant ε_{0}.
Capacitance is defined as C = Q/ΔV, where ΔV = ò_{+}^{} E•dl .
Since we know the electric field, we can determine ΔV and thus C. We choose the line integral to run along the radial direction since the electric field points in this direction. Our integral then reduces to
DV = ò_{+}^{} E•dl = ò_{R}^{R+d} E(r)dr = Q/4pe_{0}r_{R}^{R+d} = Qd/4pe_{0}R(R+d) ≈ Qd/4pe_{0}R^{2} .
So
C = Q / Qd/4pe_{0}R(R+d) = 4pe_{0}R(R+d)/d ≈ ε_{0}4πR^{2}/d .
As expected, the capacitance depends only on the geometrical factors 4πR^{2} and d and on a constant ε_{0}. Note that 4πR^{2} is the surface area of the inner sphere so C = ε_{0}A/d just like a parallel plate capacitor as long as the gap d is small.
When dealing with capacitor problems, we know the following facts:
(a) The charge on the 4 μF capacitor is
Q_{A} = CV = (4 μF)(20 V) = 80 μC .
The charge on the 8 μF capacitor is
Q_{B} = CV = (8 μF)(20 V) = 160 μC .
(b) When we also close the second switch S_{2}, the two capacitors are now connected in parallel. Capacitors in parallel must have the same voltage drop (size and direction). Here the voltage drops are opposite, so charge must flow until the voltage drop is the same.
Let's call the new charge on the capacitors Q_{L} and Q_{R}. The before and after picture of the charges on the plates looks like the diagram below.
Conservation of charge requires that
Q_{R} + Q_{L} = Q_{A}  Q_{B}  (1) 
If we apply Conservation of Energy (Kirchhoff's Rule), we find
Q_{L}/C_{L}  Q_{R}/C_{R} = 0  (2) 
Equation (2) reduces to
Q_{L} = Q_{R}(C_{L}/C_{R}) ,  (3) 
which we may substitute into equation (1) to get
Q_{R} = (Q_{A}Q_{B})[C_{R}/(C_{R}+C_{L})] = (80 μC  160 μC) [(8 μF)/(8 μF + 4 μF)] = 160/3 μF
and, recalling equation (3),
Q_{L} = (160/3 μC)(4 μF)/(8 μF) = 80/3 μC .
Alternate method:
The equivalent capacitor is
C_{AB} = C_{A} + C_{B} = 12 μF.
The total charge on the equivalent capacitor using conservation of charge is
Q_{AB} = Q_{A}  Q_{B} = 80 μC  160 μC = 80 μC ,
where we use the absolute bars to get a positive answer, and the minus sign comes from the fact that the positive plate of one capacitor was connected to the negative plate of the other.
The voltage drop over the equivalent capacitor is
V_{AB} = Q_{AB}/C_{AB} = 80 μF / 12 μC = (20/3) Volts .
Now this must also be the final voltage of each individual capacitor so
Q_{Anew} = C_{A}V_{AB} = 4 μF × 20/3 V = 80/3 μC ,
and
Q_{Bnew} = C_{B}V_{AB} = 8 μF × 20/3 V = 160/3 μC .
These are the same results we found earlier.
When we also close both switches, the two capacitors are connected in parallel. Capacitors in parallel must have the same voltage drop (size and direction). Here the voltage drops are the same, unlike the previous question, so no charge will flow.
Capacitors C_{3} and C_{4} are in series and therefore 1/C_{34} = ½ + ½ = 1, or C_{34} = 1 μF.
Capacitors C_{2} and C_{34} are in parallel and therefore C_{234} = 2 + 1 = 3 μF.
Final C_{1} and C_{234} and C_{5} are is series and therefore 1/C_{12345} = 1/3 + 1/3 +1/3 = 1, or C_{12345} = 1 μF. The equivalent capacitance is 1 μF.
The 1 μF and 3 μF capacitors are in parallel and are equivalent to a single 4 μF capacitor. The 6 μF and 2 μF capacitors are also in parallel and are equivalent to a single 8 μF capacitor.
The two 4 μF capacitors are in series, therefore 1/C_{p} = ¼ + ¼ = ½, or C_{p} = 2 μF. The two 8 F capacitors are in series, therefore 1/C_{p} = 1/8 + 1/8 = ¼, or C_{p} = 4 μF.
The 2 μF and 4 μF capacitors are in parallel and are equivalent to a single 6 μF capacitor. The equivalent capacitance is 6 μF.
Capacitance is defined as C = Q/ΔV, where ΔV = ò_{+}^{} E•dl .
We know that the presence of the dielectric reduces the electric field to E/k in the dielectric but has no other effect. Since we know the electric field everywhere, we can determine ΔV and thus C. We choose the line integral to run along the x direction since the electric field points in this direction. Our integral now reduces to a pair of integrals
DV = ò_{+}^{} E•dl = ò_{0}^{d} E(x)dx = ò_{0}^{d/3} (E/k)dx + ò_{d/3}^{d} Edx = Σd/3ke_{0} + 2Σd/3ε_{0} ,
where we take the positive plate as x = 0.
Note that Q = ΣA, so
C = ΣA / (Σd/3ke_{0} + 2Σd/3ε_{0}) = 3ε_{0}Ak/d(1 + 2k) .
Capacitance is defined as C = Q/ΔV, where ΔV = ò_{+}^{} E•dl .
We know that the presence of the dielectric reduces the electric field to E/k in the dielectric but has no other effect. Since we know the electric field, we can determine ΔV and thus C. We choose the line integral to run along the radial direction since the electric field points in this direction. Our integral then reduces to
ΔV 
= ò_{+}^{} E•dl 

= ò_{R}^{R+½d} E(r)dr + ò_{R+½d }^{R+d} E(r)dr 

= −Q/4pe_{0}r_{R}^{R+½d} − Q/4pke_{0}r_{R+½d }^{R+d} 

= ½Qd/4pe_{0}R(R+½d)+ ½Qd/4pke_{0}(R+½d)(R+d) 

≈ ½Qd/4pe_{0}R^{2} + ½Qd/4pke_{0}R^{2} 
So
C = Q / [½Qd/4pe_{0}R(R+½d)+ ½Qd/4pke _{0}(R+½d)(R+d)] ≈ ke_{0}8πR^{2 }/ d(k + 1) .
(a) The relationship between the charge density on the capacitor
plates and the field between them is E = E_{0}/k
= Σ_{free}/ke_{0}
, where E is the electric field with the dielectric in place and E_{0}
is the field when there is no dielectric. Therefore the charge density
is
Σ_{free} = Eke_{0} = (1.60 × 10^{6} V/m)(4.50)(8.85 × 10^{12}μF/m) = 63.72 μC/m^{2} .
Σ_{bound} =  [(k1)/k] Σ_{free} = 49.56 μC/m^{2} .
The minus sign indicates that the dielectric charge is opposite to the plate charge.
(a) With no dielectric, the relationship between the charge density on the capacitor plates and the field between them is E_{0} = Σ_{free}/ε_{0} , so the charge density is
Σ_{free} = E_{0}ε_{0} = (3.60 × 10^{5} V/m)(8.85 × 10^{12} μF/m) = 3.186 μC/m^{2} .
With the dielectric, the relationship between the net charge density on the capacitor plates and on the dielectric and the field between them is E = Σ_{net}/ε_{0} , so the charge density is
Σ_{net} = E_{0} = (1.20 × 10^{5} V/m)(8.85 × 10^{12} μF/m) = 1.062 μC/m^{2} .
Since the net charge density is the sum of the free and bound charge densities, we find
Σ_{bound} = Σ_{net}  Σ_{free} = 1.062 μC/m^{2}  3.186 μC/m^{2} = 2.124 μC/m^{2} .
The minus sign indicates that the dielectric charge is opposite to the plate charge.
(b) The dielectric constant is given by
k = E_{0}/E = (3.60 × 10^{5} V/m)/(1.20 × 10^{5} V/m) = 3.00 .
(a) The capacitance is given by
C_{0} = Q/V = 48 μC / 12 V = 4 μF .
(b) Since the charge on the plates has no possible path from the plate, it is unchanged. The capacitance becomes C = kC_{0} = 8.4 μF. The voltage across the capacitor will be
V = Q/C = 48 μC / 8.4 μF = 5.71 Volts.
(a) The capacitance of a parallel plate capacitor is given by
C = ke_{0}A/d = (2.1)(8.85 × 10^{12} F/m)( 1.75 × 10^{4} m^{2})/(0.04 × 10^{3} m) = 81.31 pF .
(b) From the electric field at breakdown we can find the charge density on the plates,
E = Σ_{free}/ke_{0},
or
Σ_{free} = ke_{0}E .
The voltage across the capacitor is related to the charge by
V = QΣ_{free}/C = Σ_{free}/(C/A) = dε_{free}/ke_{0} .
Combining our results yields
V = Ed = (0.04 × 10^{3} m)(60 × 10^{6} V/m) = 2400 Volts .
For each capacitor Q = CV, and we need to recall the following:
Series  Parallel 
1/C_{s} = 1/C_{1} + 1/C_{2}  C_{p} = C_{1} + C_{2} 
Same charge on C_{s}, C_{1}, and C_{2}  Same Voltage 
Different voltage  Different charge on C_{p}, C_{1}, and C_{2} 
(a) Then we work backwards from the equivalent capacitor. Since V = 600 Volts and C_{12345} = 1 μF, the charge on the equivalent capacitor is Q_{12345} = 600 μC. The energy is U_{12345} = ½QV = 0.180 J.
(b) The capacitor C_{12345} is actually C_{1}, C_{234}, and C_{5} in series, so each has the same charge of 600 μC. The voltage drop over each is V_{1} = V_{234} = V_{5} = Q/C = 600 μC / 3 μF = 200 V. The energy for each using U = ½QV yields U_{1} = U_{234} = U_{5} = 0.060 J.
(c) The capacitor C_{234} is actually C_{2} and C_{34} in parallel, so there is 200 volts over each. The charge on C_{2} is Q_{2} = C_{2}V = 400 μC. The charge on C_{34} is Q_{34} = C_{34}V = 200 μC. The energy for each using U = ½QV yields U_{2} = 0.040 J and U_{34} = 0.020 J.
(d) The capacitor C_{34} is actually C_{3} and C_{4} in series, so each has the same charge of 200 μC. The voltage drop over each is V_{2} = V_{4} = Q/C = 200 μF / 2 μF = 100 V. The energy for each using U = ½QV yields U_{2} = U_{4} = 0.010 J.
For each capacitor Q = CV, and we need to recall the following:
Series  Parallel 
1/C_{s} = 1/C_{1} + 1/C_{2}  C_{p} = C_{1} + C_{2} 
Same charge on C_{s}, C_{1}, and C_{2}  Same Voltage 
Different voltage  Different charge on C_{p}, C_{1}, and C_{2} 
(a) Then we work backwards from the equivalent capacitor. Since V = 24 Volts and C_{eq} = 6 μF, the charge on the equivalent capacitor is Q_{eq} = 24 V × 6 μF = 144 μC. The energy is U_{eq} = ½QV = 1.728 mJ.
(b) The equivalent capacitor is actually a 2 μF and a 4 μF capacitor in parallel, so each has the same voltage of 24 V. The charge on the 2 μF capacitor is Q = 2μF × 24 V = 48 μC. The charge on the 4 μF capacitor is Q = 4μF × 24 V = 96 μC. The energy for each using U = ½QV yields U_{2} = 0.576 mJ and U_{4} = 1.152 mJ.
(c) The 2 μF capacitor is actually two 4 μF capacitors in series, so each has the same charge of 48 μC. The voltage drop over each is V = Q/C = 48 μC / 4 μF = 12 V. The energy for each using U = ½QV yields U_{4R} = U_{4L} = 0.288 mJ.
The 4 μF capacitor is actually two 8 μF capacitors in series, so each has the same charge of 96 μC. The voltage drop over each is V = Q/C = 96 μC / 8 μF = 12 V. The energy for each using U = ½QV yields U_{8R} = U_{8L} = 0.576 mJ.
(d) The top right 4F capacitor is actually a 1 μF and a 3 μF capacitor in parallel, so each has the same voltage of 12 V. The charge on the 1 μF capacitor is Q = 1μF × 12 V = 12 μC. The charge on the 3 μF capacitor is Q = 3μF × 12 V = 36 μC. The energy for each using U = ½QV yields U_{1} = 0.072 mJ and U_{3} = 0.216 mJ.
The bottom left 8F capacitor is actually a 6 μF and a 2 μF capacitor in parallel, so each has the same voltage of 12 V. The charge on the 6 μF capacitor is Q = 6μF × 12 V = 72 μC. The charge on the 2 μF capacitor is Q = 2μF × 12 V = 24 μC. The energy for each using U = ½QV yields U_{6} = 0.432 mJ and U_{2} = 0.144 mJ.
(a) The presence of the dielectric gives the third capacitor a capacitance of 20 μF. Since the capacitors are in parallel, the voltage drop is 4 volts over each. The charge on each is given by Q=CV. So the charge on the top and middle capacitor is Q_{top} = Q_{middle} = 10 μF × 4 V = 40 μC. The charge on the bottom capacitor is Q_{bottom} = 20 μF × 4 V = 80 μF. The energy is given by U = ½CV^{2}. Hence the energy stored in the top and middle capacitor is U_{top} = U_{middle} = ½(10 μF)(4 V)^{2} = 80 J. The energy stored in the bottom capacitor is U_{bottom} = ½(20 μF)(4 V)^{2} = 160 J.
(b) Removing the battery changes nothing except that there is no longer an external power supply to keep the three capacitors at the same potential. Removing the dielectric changes the capacitance of the dielectric back to 10 μF. If the 80 μC stayed on the bottom capacitor, its new potential difference would be V_{bottom} = Q/C = (80 μC)/(10 μF) = 10 V. However the capacitors are in parallel they must always have the same voltage. Thus the charge will move until this occurs. Since the capacitors are now identical, the charge will drop until each capacitor has exactly the same charge. So the extra 40 μC on the bottom capacitor splits three equal ways and each capacitor will end up with a total charge of 160/3 μC. The voltage drop over each capacitor is then V = Q/C = 16/3 Volts. The energy stored in each capacitor with then be U = ½Q^{2}/C = 142.22 J.
(c) The final total energy of the three capacitors is U_{final} = 3 × 142.22 J = 426.67 J. The initial energy was U_{initial} = 80 J + 80 J + 160 J = 320 J. The capacitors gained an energy U = U_{final}  U_{initial} = 107 J. The extra energy in the system came from the person who pulled the dielectric out. The person had to do work to remove the dielectric since it had a surface charge opposite to the capacitor plates and thus was attracted to it.
Questions?mike.coombes@kwantlen.ca