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Capacitor Solutions


  1. Use the fact that the field due to a plane of charge is E = Σ/2ε0 and the Principle of Conservation of Charge to prove the following.
    (a) For very large thin conducting plate with total charge Q and area A, show that the charge and charge densities on the two sides are equal to ½Q.
    (b) Consider two very large conducting plates which are initially far apart. One has total charge Q1 and the other has total charge Q2. Show that when the plates are brought closer together so that only a small gap separates them, that the charge on each plate redistributes itself so that the sides facing the gap have equal and opposite charges and charge densities while the outer sides must have equal charges and charge densities. Determine the net electric field to the left, between, and to the right of the plates.

    (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 Q1 and Q2 on the other. The charge densities are then Σ1 = Q1/A and Σ2 = Q2/A. The electric field due to the right-side plane of charge is E1 = Σ1/2ε0 and due to the left-side plane of charge is E2 = Σ2/2ε0, 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

    Enet = E1 - E2 .

    We know that the Enet = 0 since the slab is a conductor. Thus, using our equations for the field due to a plane of charge

    0 = Σ1/2ε0 - Σ2/2ε0 .

    This yields the result that Σ1 = Σ2, so the sides have equal charge densities. Since Q = ΣA, and the sides have the same are, then Q1 = Q2 and the sides have the same charge. From Conservation of Charge, we know that Q1 + Q2 = Q, hence Q1 = Q2 = 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 ΣL = ½(Q1/A) be the initial charge density on either side of the left plate and ΣR = ½(Q2/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 DsL, 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 DsR. 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/2ε0. We assume that the field for the left plate are positive and those from the right plate are negative, we get E1 = (ΣL-DsL)/2ε0, E2 = (ΣL+DsL)/2ε0, E3 = -(ΣR+DsR)/2ε0, and E4 = -(ΣR-DsR)/2ε0, as shown in the sketch below.

    The net electric fields in the various regions are
     

    EA = -E1 + -E2 + E3 + E4 = (-2ΣL - 2ΣR)/2ε0 (1)
    EB = E1 + -E2 + E3 + E4 = (-2DsL - 2ΣR)/2ε0 (2)
    EC = E1 + E2 + E3 + E4 = (2ΣL - 2ΣR)/2ε0 (3)
    ED = E1 + E2 - E3 + E4 = (2ΣL + 2DsR)/2ε0 (4)
    EF = E1 + E2 - E3 - E4 = (2ΣL + 2ΣR)/2ε0 (5)

    The field inside each plate, EB and ED, must be zero. Therefore equation (2) and equation (4) become

    -2DsL - 2ΣR = 0 ,

    and

    L + 2DsR = 0 .

    Thus we have DssL = -ΣR and DsR = -ΣL. Thus the total charge density on each plate is ΣLR, ΣLR, ΣRL, and ΣRL 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 (ΣLR)A = ½(Q1 + Q2), (ΣLR)A = ½(Q1 - Q2), (ΣRL)A = ½(Q2 - Q1), and (ΣRL)A = ½(Q1 + Q2) 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

    Eleft = -(ΣL + ΣR)/ε0 = -(Q1 + Q2)/2Aε0 ,

    Ecentre = (2ΣL - 2ΣR)/2ε0 = (Q1 - Q2)/2Aε0 ,

    and

    Eright = +(ΣL + ΣR)/ε0 = (Q1 + Q2)/2Aε0 .

    The left and right electric fields are equal but opposite.

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  3. Two very large thin square conducting plates are very far apart. The area of the plates is A. A total charge of +Q is added to the first plate; -Q to the second. The two plates are now brought closer together so that only a small gap separates them.

  4. (a) Use the results of question 1 to find the charge on each side of each plate.
    (b) Determine the electric field
        (i)   just to the left of the plates,
        (ii)  between the plates, and
        (iii) just to the right of the plates.

    (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[Q1 - (-Q2)] = +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

    Eleft = -[Q + (-Q)]/2Aε0 = 0
    Ecentre =  [Q - (-Q)]/2Aε0 = Q/Aε0
    Eright =  [Q + (-Q)]/2Aε0 = 0

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  5. Two very large thin square conducting plates are very far apart. The area of the plates is 5.0 m2. A total charge of 7.0 μC is added to the first plate; -5.0 μC to the second. The two plates are now brought closer together so that only a small gap separates them.

  6. (a) Use the results of question 1 to find the charge on each side of each plate.
    (b) Determine the electric field
        (i)   just to the left of the plates,
        (ii)  between the plates, and
        (iii) just to the right of the plates.

    (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

    Eleft = -[(7.0 μC) + (-5.0 μC)]/2(5.0 m20 = -2.26 × 104 V/m 
    Ecentre = [(7.0 μC) - (-5.0 μC)]/2(5.0 m20 = 1.36 × 105 V/m
    Eright = [(7.0 μC) + (-5.0 μC)]/2(5.0 m20 = 2.26 × 104 V/m

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  7. Two capacitors, one with charge QA and the other with charge QB, are connected by a single wire as shown in the diagram below. Prove that the charge on the plates does not change when the switch S is closed. Use the results of questions 1 and 2.
  8. 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

    Q1 + Q2 = QA (1)
    -Q2 + Q1 + Q3 + Q4 = -QA + QB (2)
    -Q4 + Q3 = -QB (3)

    Furthermore, for the system to be in equilibrium with no charge flowing,

    Q1 = Q3 (4)

    Using equation (4) to eliminate Q3 from the first three equations we get

    Q1 + Q2 = QA (1)
    -Q2 + 2Q1 + Q4 = -QA + QB (2)
    -Q4 + Q1 = -QB (3)

    Adding equations (1), (2), and (3) together, we get

    4Q1 = 0 (4)

    So the charges on the outer plates is still zero, Q2 = QA, and Q4 = QB. Thus there has been no redistribution of charge as we had predicted.

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  9. Two capacitors are charged separately to the same potential V. They are then connected in parallel with positive plate to positive plate and negative to negative. Prove (a) that the charge on the each capacitor resides only on the inner faces of the plates and (b) that the charge on the capacitors does not change when the switch S is closed. Use the results of questions 1 and 2.

  10. 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 +QA and +QB respectively and then the upper plates would have had charge -QA and -QB. Conservation of charge says that

    -Q2 + Q1 + Q3 - Q4 = -QA - QB (1)
    Q2 + Q1 + Q3 + Q4 = QA + QB (2)

    If we add equations (1) and (2) together, we get

    2Q1 + 2Q3 = 0 (3)

    Since the capacitors must be in equilibrium, Q1 must equal Q3 if no charge is to flow. Thus equation (3) implies that

    Q1 = Q3 = 0 .

    Thus all the charge is on the inner faces and equations (2) and (3) reduce to

    Q2 + Q4 = QA + QB .

    As well, it is necessary that Q2 = QA and Q4 = QB if the voltage drop over each capacitor is to remain unchanged.

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  11. A capacitor consists of two parallel flat plates of area A with equal and opposite charge density Σ and separation d. The electric field between the plates is constant at E = Σ/ε0 and is directed from the positive plate to the negative. Find an expression for the capacitance of the parallel plate capacitor.
  12. Capacitance is defined as C = Q/ΔV, where ΔV = ò+- Edl .

    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+- Edl = ò0d Edx = Σd0 ,

    where we take the positive plate as x = 0.

    Note that Q = ΣA, so

    C = ΣA / Σd0 = ε0A /d .

    As expected, the capacitance depends only on the geometrical factors A and d and on a constant ε0.

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  13. A capacitor consists of two concentric spherical shells. The inner shell at radius R has positive charge Q. The outer shell at radius R + d has charge −Q. The distance d is much smaller than R. As a result the electric field is radial and has a magnitude E(r) = Q/4pe0r2. Find an expression for the capacitance.
  14. Capacitance is defined as C = Q/ΔV, where ΔV = ò+- Edl .

    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+- Edl = òRR+d E(r)dr = -Q/4pe0r|RR+d = Qd/4pe0R(R+d) ≈ Qd/4pe0R2 .

    So

    C = Q / Qd/4pe0R(R+d) = 4pe0R(R+d)/d ≈ ε04πR2/d .

    As expected, the capacitance depends only on the geometrical factors 4πR2 and d and on a constant ε0. Note that 4πR2 is the surface area of the inner sphere so C = ε0A/d just like a parallel plate capacitor as long as the gap d is small.

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  15. A 4.00 μF capacitor and an 8.00 μF capacitor are separately charged by a 20.0 Volt power supply. The capacitors are then placed in the circuit shown below.

  16. (a) What is the charge on each plate when both switches are open?
    (b) What will be the charge on each plate when both switches are closed?

    When dealing with capacitor problems, we know the following facts:

    1. the charge resides on the inner sides of the plates and is equal and opposite,
    2. the charge on a plate is related to the capacitance and voltage by Q = CΔV,
    3. when two capacitors are connected in series such that charge can flow (i.e. in a circuit) the capacitors end up with the same charge
    4. the equivalent capacitance of series capacitors is 1/Cseries = 1/C1 + 1/C2,
    5. when two capacitors are connected in parallel charge will flow until there is the same ΔV across each
    6. the equivalent capacitance of parallel capacitors is Cparallel = C1 + C2.
    We will use these facts to solve the following problems.

    (a) The charge on the 4 μF capacitor is

    QA = CV = (4 μF)(20 V) = 80 μC .

    The charge on the 8 μF capacitor is

    QB = CV = (8 μF)(20 V) = 160 μC .

    (b) When we also close the second switch S2, 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 QL and QR. The before and after picture of the charges on the plates looks like the diagram below.

    Conservation of charge requires that

    QR + QL = QA - QB (1)

    If we apply Conservation of Energy (Kirchhoff's Rule), we find

    QL/CL - QR/CR = 0 (2)

    Equation (2) reduces to

    QL = QR(CL/CR) , (3)

    which we may substitute into equation (1) to get

    QR = (QA-QB)[CR/(CR+CL)] = (80 μC - 160 μC) [(8 μF)/(8 μF + 4 μF)] = -160/3 μF

    and, recalling equation (3),

    QL = (-160/3 μC)(4 μF)/(8 μF) = -80/3 μC .

    Alternate method:

    The equivalent capacitor is

    CAB = CA + CB = 12 μF.

    The total charge on the equivalent capacitor using conservation of charge is

    QAB = |QA - QB| = |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

    VAB = QAB/CAB = 80 μF / 12 μC = (20/3) Volts .

    Now this must also be the final voltage of each individual capacitor so

    QAnew = CAVAB = 4 μF × 20/3 V = 80/3 μC ,

    and

    QBnew = CBVAB = 8 μF × 20/3 V = 160/3 μC .

    These are the same results we found earlier.

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  17. A 4.00 μF capacitor and an 8.00 μF capacitor are separately charged by a 20.0 Volt power supply. The capacitors are then placed in the circuit shown below. What will be the charge on each plate when both switches are closed? (Note that the polarities are switched from the previous question.)
  18. 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.

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  19. In the figure below, C1 = C5 = 3.00 μF and C2 = C3 = C4 = 2.00 μF . What is the equivalent capacitance of the circuit?

  20. Capacitors C3 and C4 are in series and therefore 1/C34 = ½ + ½ = 1, or C34 = 1 μF.

    Capacitors C2 and C34 are in parallel and therefore C234 = 2 + 1 = 3 μF.

    Final C1 and C234 and C5 are is series and therefore 1/C12345 = 1/3 + 1/3 +1/3 = 1, or C12345 = 1 μF. The equivalent capacitance is 1 μF.

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  21. What is the equivalent capacitance of the circuit shown below?
  22. 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/Cp = ¼ + ¼ = ½, or Cp = 2 μF. The two 8 F capacitors are in series, therefore 1/Cp = 1/8 + 1/8 = ¼, or Cp = 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.

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  23. A capacitor consists of two parallel flat plates of area A with equal and opposite charge density Σ and separation d. The electric field between the plates is constant at E = Σ/ε0 and is directed from the positive plate to the negative. A dielectric of thickness 1/3d and constant k is inserted next to the positive plate. Find an expression for the capacitance.

    Capacitance is defined as C = Q/ΔV, where ΔV = ò+- Edl .

    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+- Edl = ò0d E(x)dx = ò0d/3 (E/k)dx + òd/3d Edx = Σd/3ke0 + 2Σd/3ε0 ,

    where we take the positive plate as x = 0.

    Note that Q = ΣA, so

    C = ΣA / (Σd/3ke0 + 2Σd/3ε0) = 3ε0Ak/d(1 + 2k) .

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  25. A capacitor consists of two concentric spherical shells. The inner shell at radius R has positive charge Q. The outer shell at radius R + d has charge − Q. The distance d is much smaller than R. As a result the electric field is radial and has a magnitude E(r) = Q/4pe0r2. A spherical shell dielectric of radius d is inserted next to the negative shell. Find an expression for the capacitance.

    Capacitance is defined as C = Q/ΔV, where ΔV = ò+- Edl .

    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+- Edl

     

    = òRR+d E(r)dr + òR+d R+d E(r)dr

     

    = −Q/4pe0r|RR+d − Q/4pke0r|R+d R+d

     

    = Qd/4pe0R(R+d)+ Qd/4pke0(R+d)(R+d)

     

    ≈ Qd/4pe0R2 + Qd/4pke0R2

    So

    C = Q / [Qd/4pe0R(R+d)+ Qd/4pke 0(R+d)(R+d)] ≈ ke08πR2 / d(k + 1) .

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  27. Two oppositely charged conducting plates, with equal magnitude of charge per unit area, are separated by a dielectric 3.00 mm thick, with a dielectric constant of 4.50. The resultant electric field in the dielectric is 1.60 × 106 V/m. Compute (a) the charge per unit area on the conducting plates, and (b) the charge per unit area on the surfaces of the dielectric.

  28. (a) The relationship between the charge density on the capacitor plates and the field between them is E = E0/k = Σfree/ke0 , where E is the electric field with the dielectric in place and E0 is the field when there is no dielectric. Therefore the charge density is

    Σfree = Eke0 = (1.60 × 106 V/m)(4.50)(8.85 × 10-12μF/m) = 63.72 μC/m2 .

    (b) The bound charge density is related to the free charge density by

    Σbound = - [(k-1)/k] Σfree = -49.56 μC/m2 .

    The minus sign indicates that the dielectric charge is opposite to the plate charge.

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  29. Two parallel plates have equal and opposite charges. When the space between the plates is evacuated, the electric field is 3.60 × 105 V/m. When the space is filled with a dielectric, the electric field is 1.20 × 105 V/m.

  30. (a) What is the charge density on the surface of the dielectric?
    (b) What is the dielectric constant?

    (a) With no dielectric, the relationship between the charge density on the capacitor plates and the field between them is E0 = Σfree0 , so the charge density is

    Σfree = E0ε0 = (3.60 × 105 V/m)(8.85 × 10-12 μF/m) = 3.186 μC/m2 .

    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 = Σnet0 , so the charge density is

    Σnet = E0 = (1.20 × 105 V/m)(8.85 × 10-12 μF/m) = 1.062 μC/m2 .

    Since the net charge density is the sum of the free and bound charge densities, we find

    Σbound = Σnet - Σfree = 1.062 μC/m2 - 3.186 μC/m2 = -2.124 μC/m2 .

    The minus sign indicates that the dielectric charge is opposite to the plate charge.

    (b) The dielectric constant is given by

    k = E0/E = (3.60 × 105 V/m)/(1.20 × 105 V/m) = 3.00 .

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  31. A capacitor that has air between its plates is connected across a potential difference of 12 V and stores 48 μC of charge. It is then disconnected from the source while still charged. (a) find the capacitance of the capacitor. (b) A piece of Teflon (k = 2.1) is inserted between the plates. find the capacitance of, the voltage across, and charge on the capacitor.

  32. (a) The capacitance is given by

    C0 = 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 = kC0 = 8.4 μF. The voltage across the capacitor will be

    V = Q/C = 48 μC / 8.4 μF = 5.71 Volts.

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  33. Determine (a) the capacitance and (b) the maximum voltage that can be applied to a Teflon-filled (k = 2.1) parallel-plate capacitor having a plate area of 1.75 cm2 and dielectric thickness of 0.04 mm. Breakdown occurs when the electric field exceeds 60 × 106 V/m.

  34. (a) The capacitance of a parallel plate capacitor is given by

    C = ke0A/d = (2.1)(8.85 × 10-12 F/m)( 1.75 × 10-4 m2)/(0.04 × 103 m) = 81.31 pF .

    (b) From the electric field at breakdown we can find the charge density on the plates,

    E = Σfree/ke0,

    or

    Σfree = ke0E .

    The voltage across the capacitor is related to the charge by

    V = QΣfree/C = Σfree/(C/A) = dεfree/ke0 .

    Combining our results yields

    V = Ed = (0.04 × 10-3 m)(60 × 106 V/m) = 2400 Volts .

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  35. In question 10, a potential of 600 V is applied across points A and B. What is the charge on each capacitor? What is the energy stored in each capacitor?

  36. For each capacitor Q = CV, and we need to recall the following:

    Series Parallel
    1/Cs = 1/C1 + 1/C2 Cp = C1 + C2
    Same charge on Cs, C1, and C2 Same Voltage
    Different voltage Different charge on Cp, C1, and C2

    (a) Then we work backwards from the equivalent capacitor. Since V = 600 Volts and C12345 = 1 μF, the charge on the equivalent capacitor is Q12345 = 600 μC. The energy is U12345 = ½QV = 0.180 J.

    (b) The capacitor C12345 is actually C1, C234, and C5 in series, so each has the same charge of 600 μC. The voltage drop over each is V1 = V234 = V5 = Q/C = 600 μC / 3 μF = 200 V. The energy for each using U = ½QV yields U1 = U234 = U5 = 0.060 J.

    (c) The capacitor C234 is actually C2 and C34 in parallel, so there is 200 volts over each. The charge on C2 is Q2 = C2V = 400 μC. The charge on C34 is Q34 = C34V = 200 μC. The energy for each using U = ½QV yields U2 = 0.040 J and U34 = 0.020 J.

    (d) The capacitor C34 is actually C3 and C4 in series, so each has the same charge of 200 μC. The voltage drop over each is V2 = V4 = Q/C = 200 μF / 2 μF = 100 V. The energy for each using U = ½QV yields U2 = U4 = 0.010 J.

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  37. In question 11, a potential of 24 V is applied across points A and B. What is the charge on each capacitor? What is the energy stored in each capacitor?

  38. For each capacitor Q = CV, and we need to recall the following:

    Series Parallel
    1/Cs = 1/C1 + 1/C2 Cp = C1 + C2
    Same charge on Cs, C1, and C2 Same Voltage
    Different voltage Different charge on Cp, C1, and C2

    (a) Then we work backwards from the equivalent capacitor. Since V = 24 Volts and Ceq = 6 μF, the charge on the equivalent capacitor is Qeq = 24 V × 6 μF = 144 μC. The energy is Ueq = 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 U2 = 0.576 mJ and U4 = 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 U4R = U4L = 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 U8R = U8L = 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 U1 = 0.072 mJ and U3 = 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 U6 = 0.432 mJ and U2 = 0.144 mJ.

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  39. Three 10-μF capacitors are connected in parallel. A dielectric k = 2.0 is inserted into one of the capacitors. The capacitors are then connected to a 4.0 V battery. (a) What is the charge on each capacitor and what is the energy stored by each capacitor? (b) The battery is disconnected. The dielectric is then removed from the capacitor. What is the new charge on each capacitor and what now is the energy stored by each capacitor? (c) Compare the energy of the system before and after the dielectric is removed. Where did this energy come from and why?
  40. (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 Qtop = Qmiddle = 10 μF × 4 V = 40 μC. The charge on the bottom capacitor is Qbottom = 20 μF × 4 V = 80 μF. The energy is given by U = ½CV2. Hence the energy stored in the top and middle capacitor is Utop = Umiddle = ½(10 μF)(4 V)2 = 80 J. The energy stored in the bottom capacitor is Ubottom = 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 Vbottom = 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 = ½Q2/C = 142.22 J.

    (c) The final total energy of the three capacitors is Ufinal = 3 × 142.22 J = 426.67 J. The initial energy was Uinitial = 80 J + 80 J + 160 J = 320 J. The capacitors gained an energy U = Ufinal - Uinitial = 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.

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