A parallel plate capacitor has plates with an area of 2.6 m2 with a distance bet
ID: 1709252 • Letter: A
Question
A parallel plate capacitor has plates with an area of 2.6 m2 with a distance between the plates of 11.6 cm. It also has a dielectric with dielectric constant of 4.3 between its plates. It is then connected to a battery and when connected to the battery, it has a charge of 7.3 micro-Coulombs on each sheet. It is then removed from the battery and the distance between the plates is doubled (dielectric is still filling the entire space between sheets). What is the voltage between the sheets now in kV or 1 x 103 Volts?Explanation / Answer
Given that the area is A=2.6m2
the distance between the plates is d=11.6 x 10-2m
the nthe capacitance of the capacitor without dielectric is
C=oA/d
if the dielectric constant of the slab is k then
the capacitance is C1=kC
=koA/d
if q is the chargwe on the plates then the voltage is
V=q/C1
=qd/koA
=(7.3 x 10-6)(11.6 x 10-2m)/4.3 x (8.85 x 10-12)(2.6)
=8.55kV