A capacitor is constructed from two square, metal- +Q lic plates of sides l and separation d. Charges and -Q are placed on the plates, and the power sup- ply is then removed. A material of dielectric constant k is inserted a distance x into the capacitor as shown in Figure P26.78. 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 = €/2, assuming l = 5.00 cm, d = 2.00 mm, the dielectric is glass (K = 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. %3D %3D + ++Q -x- d к

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Chapter20: Electric Potential And Capacitance
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8. A capacitor is constructed from two square, metal-
lic plates of sides l and separation d. Charges +Q
and -Q are placed on the plates, and the power sup-
ply is then removed. A material of dielectric constant
K is inserted a distance x into the capacitor as shown
in Figure P26.78. 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 = €/2, assuming l = 5.00 cm, d =
2.00 mm, the dielectric is glass (K = 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.
%3D
+ ++Q
-x-
d
K
|-Q
Figure P26.78
Transcribed Image Text:8. A capacitor is constructed from two square, metal- lic plates of sides l and separation d. Charges +Q and -Q are placed on the plates, and the power sup- ply is then removed. A material of dielectric constant K is inserted a distance x into the capacitor as shown in Figure P26.78. 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 = €/2, assuming l = 5.00 cm, d = 2.00 mm, the dielectric is glass (K = 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. %3D + ++Q -x- d K |-Q Figure P26.78
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