A parallel-plate capacitor is made from two plates 12.0 cm on each side and 4.50

Question

A parallel-plate capacitor is made from two plates 12.0 cm on each side and 4.50 mm apart. Half of the space between these plates contains only air, but the other half is filled with Plexiglas of dielectric constant 3.40 (Fig. P24.72). An 18.0-V battery is connected across the plates. (a) What is the capacitance of this combination? (Hint: Can you think of this capacitor as equivalent to two capacitors in parallel?) (b) How much energy is stored in the capacitor? (c) If we remove the Plexiglas but change nothing else, how much energy will be stored in the capacitor?

Explanation / Answer

the formual of capacitance

C = e0 * A /d^2

for air

C1 = (8.854 * 10^-12) * (72 * 10^-4) / (0.0045) = 1.41664 * 10^-11 F

The other,

C2 = (8.854 * 10^-12) *(3.4) *(72 * 10^-4) /(0.0045) = 4.81 * 10^-11 F

the total capacitance of this combination is

C = C1 + C2 = 6.22 * 10^-11 F

b)

the formual for the energy

U = 0.5 * C * V^2

U = 0.5 * 6.22 * 10^-11 * 18 * 18

U = 1.00764 * 10^-8 J

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