Physics An air-filled parallel-plate capacitor has an area of 1.2 m2 and a capac
ID: 2280114 • Letter: P
Question
Physics
An air-filled parallel-plate capacitor has an area of 1.2 m2 and a capacitance of 3 nF. What is the distance between the plates? Express your answer using three significant figures. If the capacitor is connected to a 44 V battery, what is the electric field inside the plates? If the capacitor is connected to a 44 V battery, what is the magnitude of the charge on each plate? When the capacitor is connected to a 44 V battery, what is the energy stored in the capacitor? The battery is disconnected from the capacitor so that the charge stays on each plate. You push the plates inwards so that the distance between the plates is half of what it was. What is the energy stored on the plates now? How much work (positive or negative) are you doing on the plates moving them inwards? Why does this work make sense in terms of the force and displacement you apply?Explanation / Answer
1.
C=A*epsilon/d
or
3*10^-9=1.2*8.85*10^-12/d
or
d=3.54mm
2.
Efield=V/d=12429.38 N/C
3.
Q=CV=1.32*10^-7 C
4.
Energy stored=0.5*C*V^2=2.904*10^-6 Joule
5.
Energy stored=Q^2/Cnew
Cnew=A*epsilon/0.5d=6nf
hence;
Energy stored=Q^2/2Cnew=1.452*10^-6J
6.
Work done=-change in the stored energy of the capacitor
W=-(1.452*10^-6-2.904*10^-6)=1.452*10^-6J
Since the work has to be done against the electric field and hence displacement is opoosite to the field hence positive work input is required to change the configuration.............