Please help me with #20-22! Greatly appreciate some brieft explaination! The pla
ID: 1869947 • Letter: P
Question
Please help me with #20-22! Greatly appreciate some brieft explaination!
The plates of an air-filled parallel-plate capacitor each have an area of 0,.0095 m2 and are separated by a distance of 1.7 cm. A battery supplies the positively-charged plate with +5.93x1011C of charge and the negatively-charged plate with -5.93x 101IC of charge. 19. What is the capacitance of this capacitor? 0005 m2 a. 3.95x10 12 F b,) 4.95x 10'12 F C. 5.95x1012 F d. 6.95x10 1 F 0.01 20: How mueh energyis stered-in-this-eapaeiter? b. 2.56x10-103 e. 3.56x10-10 d-4.56x10-1nJ 21. The-battery remains eonneeted -to-this-eapaeiter as-the-distanee-between-the plates-is deubled. How-dees-th eapaeitanee in-this-new-eapaeiter eonfiguration-eempareto that-found-in-questien-199 a The eapacitanee remains eenstant b. The eapaeitanee doubles. e The eapaeitanee is four-times-larger. d. The eapaeitanee is-eut-in-half. The eapaeitanee is eut-te-a-qearter of the original-size 22 Hew e ees the energ -store gfew eapaeiter een hgurateneem are tot at ound-HHuesten 209 e ene es ehange: b. The-energy doubles. e The energy is four times-larger: d. The energy is-eutin-half. e The energy is eut teaquarter ef the eriginal-size.Explanation / Answer
19)
capacitance C = eo*A/d = 8.85*10^-12*0.0095/(1.7*10^-2) = 4.95*10^-12 F
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20)
energy stored E = (1/2)*Q^2/C = (1/2)*(5.93*10^-11)^2/(4.95*10^-12) = 3.56*10^-10 J
OPTION ( c)
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21)
c = eo*A/d
as the distance doubles
Cnew = eo*A/(2d) = C/2
the capacitance is cut in half
OPTION d
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22)
E = (1/2)*C*V^2
Enew = (1/2)*Cnew*V^2
as V is same
Cnew = C/2
Enew = (1/2)*(C/2)*V^2
Enew = E/2
The energy is cut in half
OPTION d