Engineering Electromagnetics, W. H. Hayt and J. A. Buck, McGraw Hill, 2001 (6th edition) 6장 족보

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Engineering Electromagnetics, W. H. Hayt and J. A. Buck, McGraw Hill, 2001 (6th edition) 연습문제 풀이 6장 풀이.

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6.11. Capacitors tend to be more expensive as their capacitance and maximum voltage, Vmax,
increase. The voltage Vmax is limited by the field strength at which the dielectric breaks
down, EBD. Which of these dielectrics will give the largest CVmax product for equal plate
areas: (a) air:
r = 1, EBD = 3 MV/m; (b) barium titanate:
r = 1200, EBD = 3 MV/m;
(c) silicon dioxide:
r = 3.78, EBD = 16 MV/m; (d) polyethylene:
r = 2.26, EBD = 4.7
MV/m? Note that Vmax = EBDd, where d is the plate separation. Also, C =
r
0A/d, and
so VmaxC =
r
0AEBD, where A is the plate area. The maximum CVmax product is found
through the maximum
rEBD product. Trying this with the given materials yields the winner,
which is barium titanate.
6.12. An air-filled parallel-plate capacitor with plate separation d and plate area A is connected to
a battery which applies a voltage V0 between plates. With the battery left connected, the
plates are moved apart to a distance of 10d. Determine by what factor each of the following
quantities changes:
a) V0: Remains the same, since the battery is left connected.
b) C: As C =
0A/d, increasing d by a factor of ten decreases C by a factor of 0.1.
c) E: We require E × d = V0, where V0 has not changed. Therefore, E has decreased by a
factor of 0.1.
d) D: As D =
0E, and since E has decreased by 0.1, D decreases by 0.1.
e) Q: Since Q = CV0, and as C is down by 0.1, Q also decreases by 0.1.
f) ρS: As Q is reduced by 0.1, ρS reduces by 0.1. This is also consistent with D having been
reduced by 0.1.
g) We: Use We = 1/2CV 2
0 , to observe its reduction by 0.1, since C is reduced by that factor.

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