A capasitor consists of two square plates, 5.6 cm on a side, separated by a 1.7

Question

A capasitor consists of two square plates, 5.6 cm on a side, separated by a 1.7 mm air gap. How much energy would be stored in the capacitor if a mica dielectric is placed between the plates? Assume the mica is 1.7 mm thick (and therefore fills the space between the plates).The charges on the plates are equal and opposite and of magnitude 270 µC. I know that it should be like this k=5 , A= 3.14x10-3 m2 , d= 1.7x10-3 m , Q= 270x10-6 c
C=KeA/d U= Q2 /2C
And I got these answers C= 8.17x10-11 F
U= 445.97 J & both are wrong answers plz help!!
A capasitor consists of two square plates, 5.6 cm on a side, separated by a 1.7 mm air gap. How much energy would be stored in the capacitor if a mica dielectric is placed between the plates? Assume the mica is 1.7 mm thick (and therefore fills the space between the plates).The charges on the plates are equal and opposite and of magnitude 270 µC. I know that it should be like this k=5 , A= 3.14x10-3 m2 , d= 1.7x10-3 m , Q= 270x10-6 c
C=KeA/d U= Q2 /2C
And I got these answers C= 8.17x10-11 F
U= 445.97 J & both are wrong answers plz help!!

Explanation / Answer

side a = 5.6 cm = 0.056 m Area A = a^ 2 = 3.136* 10^-3 m separation d = 1.7 mm= 1.7 * 10^-3 m dielectric constant of mica k = 4 magnitude of charge q = 270 µC.= 270* 10^-6 C capacitance C = KeA /d where e = permitivity of free space = 8.85 * 10^ -12 C ^ 2/ Nm^ 2 plug the values we get C = 6.53* 10^-11 F enenrgy E = q^ 2/ ( 2C ) = 558.17 J

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