A parallel-plate capacitor with vacuum between its horizontal plates has a capac

Question

A parallel-plate capacitor with vacuum between its horizontal plates has a capacitance of 26.0 mu F. A nonconducting liquid with dielectric constant 7.00 is poured into the space between the plates, filling up a fraction f of its volume. (a) Find the new capacitance as a function of f. (C_f is in mu F. Use any variable or symbol stated above as necessary.) C_f = 182f +26 (b) What should you expect the capacitance to be when f = 0? mu F Does your expression from part (a) agree with your answer? Yes No (c) What capacitance should you expect when f = 1? Does the expression from part (a) agree with your answer? Yes No

Explanation / Answer

With the liquid filling the space between the plates to height fd,the top of the fluid at the air-fluid interface develops an induceddipole layer of charge so that it acts as a thin plate with opposite charge on its upper and lower sides,thus the partially filled capacitor behaves as two capacitors in series connected at the interface. The upper and lower capacitors have separate capacitances:

Cup = eo*A/d(1-f)

Cd = keo*A/fd

k = 7

the equivalent series capacitance is

1/Cf = 1/Cup + 1/Cd

Cf = 7eoA/[7d - 7df - fd]

Cf = (eo*A/d) * (7/(7-6f)

Cf = 26(1-0.875f)^-1

part a )

f = 0

Cf = 26 uF

part c)

f = 1

Cf = 26 (1-0.875)^-1

Cf = 208 uF

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