A parallel-plate capacitor is made from two aluminum foil sheets, each 6.3 cm wi

Question

A parallel-plate capacitor is made from two aluminum foil sheets, each 6.3 cm wide and 5.4 cm long. Between the sheets is a Teflon strip of the same width and length but is 0.45cm thick. (The dielectric constant of Teflon is 2.1). The gap between the top plate and the Teflon strip is 0.35cm thick.
a) What is the capacitance of this capacitor? Hint: Is this setup similar to two capacitors wired in series or parallel? Perhaps…see notes online.
b) Find the magnitude of the charge on each plate when the capacitor is connected to a 12 V battery across pts. a and b. vaccum teflon

Explanation / Answer

A)

By C = tA/d [t is the relative permittivity of Teflon]
=>C = k x 0 x A/d [where k is the dielectric constant of Teflon & o is the permittivity of Vacuum]
=>C = [2.1 x 8.85 x 10-12 x {6.3 x 10-2 x 5.4x10-2 ]]/[.45 x 10-2 ]
=>C = 140.5 x 10-9 F

= 0.140uf

B)magnitude of the charge on each plate when the capacitor is connected to a 12 V battery across pts. a and b can be find out as follows

Charge Q coulombs = C V where C is in Farads and V volts.
and 0.140uf caps are not a manufactured value - its 0.140
12 * .14 = 1.68 microcoulombs

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