1. Refer to diagram 2. Capacitor C has square plates (sides s = 4.3 m) separated
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Question
1. Refer to diagram 2. Capacitor C has square plates (sides s = 4.3 m) separated by distance d = 8.59 mm of air. A dielectric ( = 9.12) is inserted; it has length and width s, but its height is nd, where n = 0.151; therefore it only fills 0.151 of the volume between the plates. Find the new capacitance of this arrangement, in nF HINT: Consider this to be two capacitors. How are these capacitors connected?
2. Refer to diagram 1. Capacitor C has square plates (sides s = 2.21 m) separated by distance d = 4.83 mm of air. A dielectric ( = 7.65) is inserted: it has height d and widths, but its length is ns, where n = 0.492; therefore, it fills 0.492 of the volume between the plates. Find the new capacitance of this arrangement, in nF. HINT: Consider this two be two capacitors. How are these capacitors connected?
Diagram 1 Diagram 2 nd nsExplanation / Answer
given,
length of sides = 4.3 m
distance d = 8.59 mm
dielectric k = 9.12
n = 0.151
in diagram we can see that capacitors can be seen as capacitors in series
let the capacitance of dielectric part be Cd
capacitance = e * k * Area / distance
Cd = 8.84 * 10^-12 * 9.12 * 4.3 * 4.3 / (0.151 * 8.59 * 10^-3)
Cd = 1.14924 * 10^-6 F
capacitance of remaining part
Cr = 8.84 * 10^-12 * 4.3 * 4.3 / ((1 - 0.151) * 8.59 * 10^-3)
Cr = 2.24123978 * 10^-8 F
since Cd and Cr are in series
1 / C = 1 / Cd + 1 / Cr
1 / C = 1 / (1.14924 * 10^-6) + 1 / (2.24123978 * 10^-8)
C = 2.19836737 * 10^-8 F
new capacitance of the arrangement = 2.19836737 * 10^-8 F or 21.9836737 nF
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