A pair of rectangular plates, each with an area of A = 322.6 cm2, are used to ma

Question

A pair of rectangular plates, each with an area of A = 322.6 cm2, are used to make a parallel plate capacitor with a gap d. The following capacitance vs. spacing data were measured. Draw a graph of capacitance C vs. 1/d. What is the dielectric constant, , of the capacitor? (Note that the meter may measure some extra, constant, capacitance from the wires, so your graph will not necessarily pass through zero. You can still use the slope of your line to find for the parallel plates.)

Explanation / Answer

The expression for the capacitance of a parallel plate capacitor is

C= (koA) / d

so when making a graph

y = C

x = 1/d

the slope is

m = (koA) (1)

It should make a chart

column x sets the capacitance

column b (y) sets the distance in centimeter

column c (y) calculates the inverse of b

Graph X Vs C(y)

the slope is 80.375 pF cm

X B C

C (pF) d(cm) 1/d ( 1/cm)

255.2 0.34 2.9411

137.6 0. 68 1.4705

98.4 1.02 0.9804

78.8 1.36 0.7353

67 1.70 0.5882

graphed using Excel or origin

We calculated from the expression 1

k = m /(oA)

A= 322.6 cm² = 322.6 10-4m²

m = 80.375 pF cm = 80.375 10-12 10-2 F m = 80.375 10-14 F m

k = 80.375 10-14 /(8.85 10-12 322.6 10-4 )

k = 2.815

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