12. A Leyden jar is a primitive parallel-plate capacitor that consists of a glas

Question

12. A Leyden jar is a primitive parallel-plate capacitor that consists of a glass jar with both its inside and outside surfaces covered with a conducting material. Let's say you build a Leyden jar that a 22 cm diameter base, and is coated with metal foil that goes up to a height of 35.5 cm. The glass has a thickness of 2.25 mm and a dielectric constant of 7.2. Estimate its capacitance 13. A 7.7 F capacitor is charged by a 125 V battery and then is disconnected from the battery When this capacitor Ci is then connected to a second (initially uncharged) capacitor Ca , the voltage on the first drops to 15 V. What is the value of Ca? doubled. By what factor does the electric potential energy stored in the capacitor change? is larger by a factor of five. If the charge on the capacitor is held fixed, by what factor does the 14. A parallel plate capacitor has fixed charges+Q and -Q. The separation of the plates is then 15. The dielectric in a parallel-plate capacitor is changed to a material with a dielectric constant that energy stored in the capacitor change? Explain why the energy is different in the two cases.

Explanation / Answer

12. for the leyden jar

nase diameter, d = 22 cm

h = 35.5 cm

t = 2.25 mm

k' = 7.2

cpaacitance C = 2*pi*epsilon*k'*h/ln((d/2 + t)/(d/2))

C = 7.2*0.355/2*8.98*10^9*ln((0.11+2.25*10^(-3))/0.11)

C = 7.02860*10^-9 F

13. C = 7.7 uF

Vi = 125 V

Vf = 15 V

now,

intiial charge, Q = CVi = 962.5uC

final charge = q

now,

q/C1 = (Q - q)/'C2

hence

C2 = (Q - q)C1/q = (Q/q - 1)C1

also

q/C1 = 15

q = 15C1

Q = 125C1

hence

C2 = (125/15 - 1)C1

C2 = 7058.33333 uF

14. initial charge = Q

initial distance = d

initial potential difference = V

intiial energy = 0.5CV^2

final capacitance = C' = C/2

final voltage remains the same = V

hence

final energy = 0.5Q^2/C

hecne potential energy increases by the factor of 2

15, charge on capacitor is constant

dielectric constant is increased by a factor of 5

so, capacitance increases by factor of 5

for same charge, E = 0.5Q^2/C

so energy stored decreases by a factor of 5

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