A parallel-plate capacitor with only air between the plates is charged by connecting it to a battery. The capacitor is then disconnected from the battery, without any of the charge leaving the plates.(a) A voltmeter reads 45.0 V when placed across the capacitor. When a dielectric is inserted between the plates, completely filling the space, the voltmeter reads 11.5 V. What is the dielectric constant of this material?
(b) What will the voltmeter read if the dielectric is now pulled partway out so it fills only one-third of the space between the plates?
Answers
(A) 3.9
When a dielectric is inserted between the plates of a capacitor, the capacitance of the capacitor increases according to the equation:
(1)
where
C' is the final capacitance
k is the dielectric constant
C is the original capacitance
The capacitance is inversely proportional to the to voltage across the plates:
(2a)
where Q is the charge stored and V the potential difference across the plates. We can rewrite C' (the capacitance of the capacitor filled with dielectric) as
(2b)
Substituting (2a) and (2b) into (1), we find
(3)
where
V = 45.0 V is the original voltage across the capacitor
V' = 11.5 V is the voltage across the capacitor filled with dielectric
Solving for k,
(B) 22.8 V
When the dielectric is partially pulled away, the system can be assimilated to a system of 2 capacitors in parallel, of which one of them is filled with dielectric and the other one is not.
Keeping in mind that the capacitance of a parallel-plate capacitor is proportional to the area of the plates:
and in this case, the area of the capacitor filled with dielectric is just 1/3 of the total, we can write:
where C1 is the capacitance of the part non-filled with dielectric, and C2 is the capacitance of the part filled with dielectric. The total capacitance of the system in parallel is
Substituting,
This is equivalent to a capacitor completely filled with a dielectric with dielectric constant k=1.97. Therefore, using again eq.(3), we find the new voltage:
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