An electronics technician wishes to construct a parallelplate capacitor using rutile (k=100) as the dielectric. The area of the

Question

2 years ago

11

plates is 1.00cm² . What is the capacitance if the rutile thickness is 1.00mm? (a) 88.5 pF (b) 177 pF (c) 8.85 μF (d) 100 μF (e) 35.4μF

1 answer:

My name is Ann [436] · Answer 1 · 2023-11-11T03:27:40+03:00

Given k=100, Area if plate A = 1.00 cm2

$= 1 x (10-2)2 m^{2}$

Thickness of rutile $ds = t = 1.00 mm$

$= 1 x 10-3 cm$

Without any dielectric medium capacitance of a parallel plate capacitor is c = oAd

When a dielectric is placed between capacitors plates then capacitance c1 = KoAd

Here $d = t = 1 x 10-1 cm$ , $o = 8.854 x 10-12 c^{2} m^{-2} N$

Now $c1 = 100 X 8.859 X 10^{-12} X 1 X (10-2)^{21} X 10^{-2}$

$= 8.859 X 10-1910-3= 8.85 X 10-11 F= 8.85 X 10-12 F$

$C1 = 88.5 X 10-12 F$

$C1 = 88.5 PF$

Correct Option: A

What purposes serve parallel plate capacitors?

The following are some uses for parallel plate capacitors:

This kind of capacitor is utilized in rechargeable energy systems such as batteries.
These capacitors are used in systems for dynamic digital memory.
Such capacitors are used in pulsed laser and radar circuits.

To learn more about Parallel plate capacitors, visit:

#SPJ4