Question:Please show all work. A parallel-plate capacitor is made from two plate

Question

Question:Please show all work.

A parallel-plate capacitor is made from two plates 10.2cm on each side and 5.61mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.4. (See the figure below(Figure 1).) An 18.09V battery is connected across the plates. What is the capacitance of this combination? How much energy is stored in the capacitor? If we remove the Plexiglas, but change nothing else, how much energy will be stored in the capacitor?

Explanation / Answer

C1 = 8.854*10^-12*0.102^2/(4*0.00561) = 4.1*10^-12 F

C2 = 8.854*10^-12*3.4*0.102^2/(4*0.00561) = 1.396*10^-11 F

So, Cnet = C1+C2 = 1.81*10^-11 F <----answer

energy stored = 0.5*Cnet*V^2 = 0.5*(1.81*10^-11)*18.09^2 = 2.96*10^-9 J

enrgy store after removal = 0.5*2*4.1*10^-12*18.09^2 = 1.34*10^-9 J

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