A parallel plate capacitor with an air gap is connected to a 6V battery. After c
ID: 1438230 • Letter: A
Question
A parallel plate capacitor with an air gap is connected to a 6V battery. After charging, the energy stored in the capacitor is 72nJ. Without disconnecting the capacitor from the battery, a dielectric material is inserted into the air gap and the energy stored in the capacitor is increased to 317nJ. Each of the plates has an area of 50 cm^2. (a) What is the charge on the positive plate of the capacitor after the dielectric is inserted? (b) What is the magnitude of the electric field between the plates before the dielectric is inserted? (c) What is the dielectric constant of the material?Explanation / Answer
(a)Let initial capacitance be C F
energy=0.5*C*V^2=72*10^(-9)
C=4nF
Charge=CV=24 nC
(b)electric field=V/d
C=epsilon0*area/distance
distance=d=1.106*10^(-5)
E=V/d=542127.85 N/C
(c) after dielectric is placed
new capacitance=Cnew =k*C=4k nF
new energy=0.5*V^2*Cnew=317 nJ
0.5*36*(4k nF)=317nJ
k=4.4