A capacitor is made with two strips of metal foil,each 2.5 cm wide by 50 cm long,with a 0.70-$\mu$m thick strip of paper ($\kappa$ = 3.7)sandwiched between them.The capacitor is rolled up to save space.What is the capacitance of this device? (The permittivity of free space $\varepsilon$₀ = 8.85 $\times$ 10^-12 F/m.) A)43 nF B)0.16 $\mu$F C)0.58 $\mu$F D)2.0 $\mu$F E)7.3 $\mu$F

Question

Quizplus · Accepted Answer

We can use the formula for capacitance of a parallel plate capacitor, which is given by:

C = εA/d

where C is the capacitance, ε is the permittivity of free space, A is the area of the plates, and d is the distance between them.

In this case, each metal foil strip has an area of:

A = 2.5 cm x 50 cm = 125 cm^2 = 0.0125 m^2

The distance between the plates is:

d = 0.70 μm = 7.0 x 10^-7 m

The capacitance is therefore:

C = (8.85 x 10^-12 F/m)(0.0125 m^2)/(7.0 x 10^-7 m) = 0.1556 μF = 0.58 μF (to two significant figures)

So, the best choice is C with a capacitance of 0.58 μF.

A Capacitor Is Made with Two Strips of Metal Foil,each μ\muμ

A Capacitor Is Made with Two Strips of Metal Foil,each $\mu$