Question

A parallel plate capacitor consists of two metal plates, each of area A = 0.015 m^2 , separated by a vacuum gap d = 0.06 m thick. What is the capacitance of this device? What voltage difference must be applied between the plates if the capacitor is to hold a charge of magnitude Q = 1.0*10^(-9) C on each plate? (dielectric constant = 1)

a. 2.2*10^(-12) F and 203 V

b. 8.8*10^(-11) F and 363 V

c. 2.2*10^(-12) F and 452 V
   
d. 5.2*10^(-12) F and 851 V

Answer 1

capacitance of parallel plate capacitor

C=A0k/d

where A=area of each plate=0.015m2

0=permitivity of vacuum=8.85x10-12

k=dielectric constat of vacuum=1

d=distance between plates=0.06m

C=A0k/d= 0.015x8.85x10-12x1/0.06=2.2x10-12F

Voltage difference between plates

V=Q/C (beacuse Capacitance C=Q/V)

where Q=charge on plate=1x10-9C, C=2.2x10-12F

V=1x10-9/2.21310-12=452V