General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 16, Problem 59E

(a)

To determine

The charge on the plates.

(a)

Expert Solution
Check Mark

Answer to Problem 59E

The charge on the plates is 8.85×108C.

Explanation of Solution

Write the expression for the capacitance.

  C=Aε0l        (I)

Here, ε0 is the permittivity of free space, A is the area, and l is the distance between the capacitors.

Write the expression for the capacitance.

  Q=CV        (II)

Here, Q is the charge and V is the potential difference.

Solve (I) and (II) for Q,

Q=(Aε0l)V        (III)

Conclusion:

Substitute 1000V for V, and 0.01m for l, 0.1m2 for A, and 8.854×1012C2/Nm2 for ε0 in equation (III) to find Q.

  Q=(1000V)(0.1m2)(8.854×1012C2/Nm2)(0.01m)=8.85×108C

Therefore, charge on the plates is 8.85×108C.

(b)

To determine

The energy stored in the capacitor.

(b)

Expert Solution
Check Mark

Answer to Problem 59E

The energy stored in the capacitor is 4.42×105.

Explanation of Solution

Write the expression for the energy stored in the capacitor.

  U=12CV2        (IV)

Here, U is the energy stored in the capacitor.

Solve for U using (IV) and (I).

  U=12(Aε0l)V2        (V)

Conclusion:

Substitute 1000V for V, and 0.01m for l, 0.1m2 for A, and 8.854×1012C2/Nm2 for ε0 in equation (V) to find U.

  U=12((0.1m2)(8.854×1012C2/Nm2)0.01m)(1000V)2=4.42×105J

Therefore, the energy stored in the capacitor is 4.42×105

(c)

To determine

The energy and potential difference of the capacitor under the insertion of dielectric slab.

(c)

Expert Solution
Check Mark

Answer to Problem 59E

The energy and potential difference of the capacitor under the insertion of dielectric slab are 4.42×106J_ and 100V_.

Explanation of Solution

Write the expression for the capacitance under the insertion of dielectric slab is,

  CK=Kε0Al        (VI)

Here, CK is the capacitance under the insertion of dielectric slab and K is the dielectric constant.

Write the potential difference for the capacitance under the insertion of dielectric slab.

  VK=QCK        (VII)

Here, VK is the potential difference for the capacitance under the insertion of dielectric slab.

Write the expression for the energy stored in the capacitor under the insertion of dielectric slab.

  UK=12CKVK2        (VIII)

Here, UK is the energy stored in the capacitor under the insertion of dielectric slab.

Conclusion:

Substitute 10 for K, and 0.01m for l, 0.1m2 for A, and 8.854×1012C2/Nm2 for ε0 in equation (VI) to find CK.

  CK=(10)(8.854×1012C2/Nm2)(0.1m2)0.01m=8.85×1010F

Substitute 8.85×108C for Q and 8.85×1010F for CK in equation (VII) to find VK.

  VK=8.85×108C8.85×1010F=100V

Substitute 100V for VK and 8.85×1010F for CK in equation (VIII) to find UK.

  U=12(8.85×1010F)(100V)2=4.42×106J

Therefore, the energy and potential difference of the capacitor under the insertion of dielectric slab are 4.42×106J_ and 100V_.

(d)

To determine

The explanation for the change.

(d)

Expert Solution
Check Mark

Answer to Problem 59E

The energy stored reduces by the factor K due to the insertion of dielectric constant.

Explanation of Solution

Rearrange the equation (II) to get V.

  V=QC        (IX)

Write the energy stored in the capacitor of medium dielectric constant.

U=12C(QC)2=Q22C

  U=Q22C        (X)

Write the energy stored in the capacitor of medium dielectric constant using (X).

  UK=Q22KC0        (XI)

Here, C0 is the dielectric constant of air.

Conclusion:

Solve the equation (X) and (XI).

U=UKK

Therefore, the energy stored reduces by the factor K due to the insertion of dielectric constant.

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Chapter 16 Solutions

General Physics, 2nd Edition

Ch. 16 - Prob. 11RQCh. 16 - Prob. 12RQCh. 16 - Prob. 13RQCh. 16 - Prob. 1ECh. 16 - Prob. 2ECh. 16 - Prob. 3ECh. 16 - Prob. 4ECh. 16 - Prob. 5ECh. 16 - Prob. 6ECh. 16 - Prob. 7ECh. 16 - Prob. 8ECh. 16 - Prob. 9ECh. 16 - Prob. 10ECh. 16 - Prob. 11ECh. 16 - Prob. 12ECh. 16 - Prob. 13ECh. 16 - Prob. 14ECh. 16 - Prob. 15ECh. 16 - Prob. 16ECh. 16 - Prob. 17ECh. 16 - Prob. 18ECh. 16 - Prob. 19ECh. 16 - Prob. 20ECh. 16 - Prob. 21ECh. 16 - Prob. 22ECh. 16 - Prob. 23ECh. 16 - Prob. 24ECh. 16 - Prob. 25ECh. 16 - Prob. 26ECh. 16 - Prob. 27ECh. 16 - Prob. 28ECh. 16 - Prob. 29ECh. 16 - Prob. 30ECh. 16 - Prob. 31ECh. 16 - Prob. 32ECh. 16 - Prob. 33ECh. 16 - Prob. 34ECh. 16 - Prob. 35ECh. 16 - Prob. 36ECh. 16 - Prob. 37ECh. 16 - Prob. 38ECh. 16 - Prob. 39ECh. 16 - Prob. 40ECh. 16 - Prob. 41ECh. 16 - Prob. 42ECh. 16 - Prob. 43ECh. 16 - Prob. 44ECh. 16 - Prob. 45ECh. 16 - Prob. 46ECh. 16 - Prob. 47ECh. 16 - Prob. 48ECh. 16 - Prob. 49ECh. 16 - Prob. 50ECh. 16 - Prob. 51ECh. 16 - Prob. 52ECh. 16 - Prob. 53ECh. 16 - Prob. 54ECh. 16 - Prob. 55ECh. 16 - Prob. 56ECh. 16 - Prob. 57ECh. 16 - Prob. 58ECh. 16 - Prob. 59ECh. 16 - Prob. 60ECh. 16 - Prob. 61ECh. 16 - Prob. 62ECh. 16 - Prob. 63ECh. 16 - Prob. 64ECh. 16 - Prob. 65ECh. 16 - Prob. 66ECh. 16 - Prob. 67ECh. 16 - Prob. 68ECh. 16 - Prob. 69ECh. 16 - Prob. 70ECh. 16 - Prob. 72ECh. 16 - Prob. 73ECh. 16 - Prob. 74ECh. 16 - Prob. 75ECh. 16 - Prob. 76ECh. 16 - Prob. 78ECh. 16 - Prob. 81ECh. 16 - Prob. 82ECh. 16 - Prob. 83ECh. 16 - Prob. 84ECh. 16 - Prob. 85ECh. 16 - Prob. 86ECh. 16 - Prob. 87ECh. 16 - Prob. 88ECh. 16 - Prob. 89ECh. 16 - Prob. 90ECh. 16 - Prob. 91ECh. 16 - Prob. 92ECh. 16 - Prob. 93ECh. 16 - Prob. 94ECh. 16 - Prob. 95ECh. 16 - Prob. 96ECh. 16 - Prob. 97ECh. 16 - Prob. 98ECh. 16 - Prob. 99ECh. 16 - Prob. 100ECh. 16 - Prob. 101ECh. 16 - Prob. 102ECh. 16 - Prob. 103ECh. 16 - Prob. 104ECh. 16 - Prob. 105ECh. 16 - Prob. 106ECh. 16 - Prob. 107ECh. 16 - Prob. 108E
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