
Concept explainers
A parallel-plate capacitor has square plates of side s = 2.50 cm and plate separation d = 2.50 mm. The capacitor is charged by a battery to a charge Q = 4.00 μC, after which the battery is disconnected. A porcelain dielectric (κ = 6.5) is then inserted a distance y = 1.00 cm into the capacitor (Fig. P27.88). Hint: Consider the system as two capacitors connected in parallel. a. What is the effective capacitance of this capacitor? b. How much energy is stored in the capacitor? c. What are the magnitude and direction of the force exerted on the dielectric by the plates of the capacitor?
Figure P27.88
(a)

Effective capacitance of the capacitor.
Answer to Problem 88PQ
Effective capacitance of the capacitor is
Explanation of Solution
Write the expression to find the capacitance of the parallel plate capacitor with dielectric.
Here,
Substitute
Write the expression to find the capacitance of the parallel plate which is unfilled.
Here,
Substitute
The system can be considered as two capacitors are connected in parallel.
Write the expression to find the equivalent capacitance of the two capacitors connected in parallel.
Here,
Conclusion:
Substitute
Therefore, effective capacitance of the capacitor is
(b)

The energy stored in the capacitor.
Answer to Problem 88PQ
The energy stored in the capacitor is
Explanation of Solution
Write the expression to find the equivalent capacitance of the two capacitors connected in parallel.
Here,
Substitute equations (I) and (II) in above expression to find the equivalent capacitance
Write the equation for energy stored in the capacitor.
Here,
Substitute equation (III) in above expression to find the energy stored in the capacitor.
Conclusion:
Substitute
Therefore, the energy stored in the capacitor is
(c)

The magnitude and direction of the force exerted on dielectric by the plates of capacitor.
Answer to Problem 88PQ
The magnitude and direction of the force exerted on dielectric by the plates of capacitor is
Explanation of Solution
Write the equation for energy stored in the capacitor.
Here,
Substitute equation (III) in above expression to find the energy stored in the capacitor.
Write the expression to find the force exerted on dielectric by the plates of capacitor.
Substitute equation (IV) in above expression to find the force exerted on dielectric by the plates of capacitor.
Conclusion:
Substitute
Therefore, the magnitude and direction of the force exerted on dielectric by the plates of capacitor is
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Chapter 27 Solutions
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