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Unformatted text preview: ially cancels the electric field
produced by the charges on the capacitor plates.
IDENTIFY:! Capacitance depends on geometry, and the introduction of a dielectric increases the capacitance.
SET UP:! For a parallel-plate capacitor, C # K !0 A/d .
EXECUTE:! (a) Solving for d gives
d# K !0 A (3.0)(8.85 " 10 !12 C 2 /N + m 2 )(0.22 m)(0.28 m)
# 1.64 " 10!3 m = 1.64 mm .
1.0 " 10!9 F 1.64 mm
C 8 sheets .
(1.0 " 10!9 F)(0.012 m)
(b) Solving for the area of the plates gives A #
# 0.45 m 2 .
K !0 (3.0)(8.85 " 10!12 C2 /N + m 2 )
(c) Teflon has a smaller dielectric constant (2.1) than the posterboard, so she will need more area to achieve the same
EVALUATE:! The use of dielectric makes it possible to construct reasonable-sized capacitors since the dielectric
increases the capacitance by a factor of K.
IDENTIFY and SET UP:! For a parallel-plate capacitor with a dielectric we can use the equation C # K !0 A / d . Dividing this result by the thickness of a sheet of paper gives 24.41. 24.42. 24.43. Minimum A means smallest possible d. d is limited by the requirement that E be less than 1.60 " 107 V/m when V is
as large as 5500 V.
EXECUTE:! V # Ed so d # #
# 3.44 " 10!4 m
E 1.60 " 107 V/m
(1.25 " 10!9 F)(3.44 " 10!4 m)
Then A #
# 0.0135 m 2 .
K !0 (3.60)(8.854 " 10!12 C2 / N + m2 )
EVALUATE:! The relation V = Ed applies with or without a dielectric present. A would have to be larger if there were
IDENTIFY and SET UP:! Adapt the derivation of Eq.(24.1) to the situation where a dielectric is present.
EXECUTE:! Placing a dielectric between the plates just results in the replacement of ! for !0 in the derivation of
Equation (24.20). One can follow exactly the procedure as shown for Equation (24.11).
EVALUATE:! The presence of the dielectric increases the energy density for a given electric field.
IDENTIFY:! The permittivity ! of a material is related to its dielectric con...
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This document was uploaded on 03/11/2014 for the course PHYSICS 240 at University of Michigan.
- Fall '08