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A parallel-plate capacitor is made from two plates 12.0 cm

Chapter 24, Problem 72P

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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. P24.66). 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?

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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. P24.66). 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?

ANSWER:

Solution 72P 0 The expression for capacitance of a parallel plate capacitor is given by C = d , A is the area of each plate and d is the separation between the plates. 4 2 The area of each plate = 12.0 cm × 12.0 cm = 144 × 10 m Area of the plexiglass portion = (144 × 10 4/2 m = 72 × 10 4 m 2 2 2 Area of the air portion = 72 × 10 m Separation of the plates d = 4.50 mm = 4.50 × 10 3 m 0A 8.85×1012 4 (a) ow, capacitance due to plexiglass C = 1 d = 3.

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