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A parallel-plate capacitor consistsof plates of area 1.5 x 10^-4  m^2. And separated...

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lkehoe | (Level 1) Valedictorian

Posted April 30, 2013 at 5:10 AM via web

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A parallel-plate capacitor consistsof plates of area 1.5 x 10^-4  m^2. And separated by 1.0 mm.  The capacitor is connected to a 12 -V battery.  (a) What is the capacitance?  (b) What is the charge on the plates? (c)  How much energy is stored in the capacitor?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted April 30, 2013 at 7:01 AM (Answer #1)

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The capacitance of a parallel plate capacitance is given by `C = (k*epsi_o*A)/d` where `epsi_o` is the permittivity of free space and equal to `8.854*10^-12` F/m, k is the relative permittivity of the dielectric material between the plates, A is the area of the plates and d is the distance by which they are separated.

The capacitor in the problem consists of plates of area 1.5 x 10^-4  m^2, separated by 1.0 mm. It is assumed the material between them is air.

C = `(1.5*10^-4*8.854*10^-12*1)/(10^-3)` = 1.3281*10^-12 F

When the capacitor is connected to a voltage source of 12 V, the charge on the plates is 1.3281*10^-12*12 = 1.5937*10^-11 C

The energy stored in the the capacitor is `(1/2)*C*V^2` = (1/2)*1.3281*10^-12*144 = 9.5623*10^-11 J

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