as you can see, i have already found the answer for 1 and 2, i just need the ans
ID: 2273089 • Letter: A
Question
as you can see, i have already found the answer for 1 and 2, i just need the answers for 3,4 and 5. please be careful with the units(u can see what units is required for each section)
Two parallel plates, each having area A = 3540cm2 are connected to the terminals of a battery of voltage Vb = 6 V as shown. The plates are separated by a distance d = 0.52cm. You may assume (contrary to the drawing) that the separation between the plates is small compared to a linear dimension of the plate. What is C, the capacitance of this parallel plate capacitor? mu F What is Q the charge stored on the top plate of the this capacitor?. mu C A dielectric having dielectric constant k = 4 is now inserted in between the plates of the capacitor as shown. The dielectric has area A = 3540 cm2 and thickness equal to half of the separation (= 0.26 cm). What is the charge on the top plate of this capacitor? mu C What is U, the energy stored in this capacitor? J The battery is now disconnected from the capacitor and then the dielectric is withdrawn. What is V, the voltage across the capacitor? VExplanation / Answer
1.
C=eoA/d
C=(8.85*10^-12)*(3540*10^-4)/(0.52*10^-2)
C=6.02*10^-10 F
C =0.000602 uF
2.
Q=CV =6*0.000602
Q=0.0036 uC
3.
C=KeoA/d
C=4*(8.85*10^-12)*(3540*10^-4)/(0.26*10^-12)
C=4.82*10^-9 F
Q=CV =4.82*10^-9*6
Q=0.0289 uC
4)
U=(1/2)CV^2=(1/2)*(4.82*10^-9)*6^2
U=8.676*10^-8 C
5)
Q stays the same and V changes
V=Q/C =8.676*10^-8/4.82*10^-9
V=18 V
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