Concept explainers
A capacitor is constructed from two square, metallic plates of sides ℓ and separation d. Charges +Q and −Q are placed on the plates, and the power supply is then removed. A material of dielectric constant κ is inserted a distance x into the capacitor as shown in Figure P25.49 (page 690). Assume d is much smaller than x. (a) Find the equivalent capacitance of the device. (b) Calculate the energy stored in the capacitor. (c) Find the direction and magnitude of the force exerted by the plates on the dielectric. (d) Obtain a numerical value for the force when x = ℓ/2, assuming ℓ = 5.00 cm, d = 2.0 mm, the dielectric is glass (κ = 4.50), and the capacitor was charged to 2.00 × 103 V before the dielectric was inserted. Suggestion: The system can be considered as two capacitors connected in parallel.
Figure P25.49
(a)
Answer to Problem 26.78CP
Explanation of Solution
Given Info: The sides of the metallic plate is
Explanation:
The given Figure of the system is shown below.
Figure (1)
The system can be considered as two capacitors one with a dielectric and one without dielectric. The top plates of the two capacitors always have same potential
Formula to calculate the area of the with dielectric is,
Here,
Formula to calculate the capacitance with dielectric is,
Here,
Substitute
Formula to calculate the area of the without dielectric is,
Here,
Formula to calculate the capacitance without dielectric is,
Here,
Substitute
Since, the capacitors
Substitute
Conclusion:
Therefore, the equivalent capacitance of the device is
(b)
Answer to Problem 26.78CP
Explanation of Solution
Given Info: The sides of the metallic plate is
Explanation:
The equivalent capacitance of the device is,
Formula to calculate the energy stored in the capacitor is,
Substitute
Conclusion:
Therefore, the energy stored in the capacitor is
(c)
Answer to Problem 26.78CP
Explanation of Solution
Given Info: The sides of the metallic plate is
Explanation:
As, the dielectric slab is inserted the potential energy of the system is affected. The
Formula to calculate the force exerted by the plates on the dielectric is,
Here,
Substitute
The negative value of the force indicates the dielectric is pushed out from the capacitor.
Conclusion:
Therefore, the magnitude of the force exerted by the plates on the dielectric is
(d)
Answer to Problem 26.78CP
Explanation of Solution
Given Info: The sides of the metallic plate is
Explanation:
The distance of plate in the dielectric medium is,
Formula to calculate the charge is,
Here,
The magnitude of the force exerted by the plates on the dielectric is,
Here,
Substitute
Substitute
Substitute
Conclusion:
Therefore, the magnitude of the force exerted by the plates on the dielectric is
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Chapter 26 Solutions
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
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