General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Question
Chapter 16, Problem 54E

(a)

To determine

The charges on the plates of the capacitor.

(a)

Expert Solution
Check Mark

Answer to Problem 54E

The charges on the plates is 1.0×104C_.

Explanation of Solution

Write the expression for the capacitance.

  Q=CV        (I)

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

Conclusion:

Substitute 2μF for C and 50V for V in equation (I) to find Q.

  Q=(2μF×1F1×106μF)(50V)=1.0×104C

Therefore, the charges on the plates is 1.0×104C_.

(b)

To determine

The electric field between the plates of the capacitor.

(b)

Expert Solution
Check Mark

Answer to Problem 54E

The electric field between the plates is 50000Vm1_.

Explanation of Solution

Write the expression for the electric field is,

  E=Vl        (II)

Here, V is the potential difference, E is the electric field, and l is the distance between the plates of the capacitor.

Conclusion:

Substitute 50V for V and 103m for l in equation (I) to find V.

  E=50V103m=50000Vm1

Therefore, the electric field between the plates is 50000Vm1_.

(c)

To determine

The new charge when dielectric constant inserted between the plates.

(c)

Expert Solution
Check Mark

Answer to Problem 54E

The new charge when dielectric constant inserted between the plates is 2.0×104C_.

Explanation of Solution

Write the expression for the new charge when dielectric constant inserted between the plates is,

  Q'=KCV        (III)

Here, Q' is the new charge when dielectric constant inserted between the plates of the capacitor, and K is the dielectric field.

Conclusion:

Substitute 2μF for C, 5 for K, and 50V for V in equation (I) to find Q.

  Q'=(5)(2μF×1F1×106μF)(50V)=2.0×104C

Therefore, the new charge when dielectric constant inserted between the plates is 2.0×104C_.

(d)

To determine

The electric field between the plates of the capacitor on insertion of dielectric.

(d)

Expert Solution
Check Mark

Answer to Problem 54E

The electric field between the plates of the capacitor on insertion of dielectric is 1.0×105Vm1_.

Explanation of Solution

Write the expression for the new potential difference when dielectric constant inserted between the plates is,

  V'=Q'C        (IV)

Here, V' is the new potential difference when dielectric constant inserted between the plates.

Write the expression for the electric field between the plates of the capacitor on insertion of dielectric is,

  E'=V'l        (V)

Here, E' is the electric field between the plates of the capacitor on insertion of dielectric.

Substitute equation (IV) in (V) to get E',

  E'=Q'Cl        (VI)

Conclusion:

Substitute 2.0×104C for Q', 103m for l, and 2μF for C in equation (VI) to find V'.

  E'=2.0×104C(2μF×1F1×106μF)103m=1.0×105Vm1

Therefore, the electric field between the plates of the capacitor on insertion of dielectric is 1.0×105Vm1_.

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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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