Appreciate help with this homework problem. Please don\'t skipsteps so I can und

Question

Appreciate help with this homework problem. Please don't skipsteps so I can understand better. Thank you. A 2.7 F parallel-plate capacitor with a distance of .26 mmbetween the plates is to be built.

(a) What is the required area of the plates without adielectric between them?
1 m2
(b) What is the required plate area if, instead, the plates arefilled with polystyrene ( = 2.6) .26 mm thickbetween them?
2 m2

Appreciate help with this homework problem. Please don't skipsteps so I can understand better. Thank you. A 2.7 F parallel-plate capacitor with a distance of .26 mmbetween the plates is to be built.

(a) What is the required area of the plates without adielectric between them?
1 m2
(b) What is the required plate area if, instead, the plates arefilled with polystyrene ( = 2.6) .26 mm thickbetween them?
2 m2

Explanation / Answer

The capacitance of the parallel plate capacitor is C = 2.7F The distance between the plates of the parallel-platecapacitor is d = .26 mm = .26 * 10-3 m (a)The capacitance of the parallel plate capacitor is C = (oA/d) or A = (C * d/o) Here,o= 8.85 * 10-12C2/Nm2 Substituting the values in the above equation,we get A = (2.7 * .26 * 10-3/8.85 *10-12) or A = 0.0793 * 109 = 7.93 * 107m2 (b)The dielectric constant of polystyrene is k = 2.6 The thickness of polystyrene between the plates is d = .26 mm= .26 * 10-3 m Therefore,the required plate area is A = (C * d/k * o) or A = (2.7 * .26 * 10-3/2.6 * 8.85 *10-12) or A = 0.0305 * 109 m2 = 3.05 *107 m2 (b)The dielectric constant of polystyrene is k = 2.6 The thickness of polystyrene between the plates is d = .26 mm= .26 * 10-3 m Therefore,the required plate area is A = (C * d/k * o) or A = (2.7 * .26 * 10-3/2.6 * 8.85 *10-12) or A = 0.0305 * 109 m2 = 3.05 *107 m2

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