A parallel-plate capacitor is made from two plates each with 10 cm diameter, spa

Question

A parallel-plate capacitor is made from two plates each with 10 cm diameter, spaced. 700 cm apart. It is connected to a 1 10 V battery. What is the capacitance? What is the charge of the capacitor? What is the electric field between the plates? What is the energy stored in the capacitor? If the battery is disconnected and then the space between the two plates is filled with dielectric material with k = 2.5, What is the capacitance? What is the charge of the capacitor? What is the electric field between the plates? What is the energy stored in the capacitor?

Explanation / Answer

Capacitance, C = e0 A / d

C = (8.854 x 10^-12)(pi (10 x 10^-2 / 2)^2) / (0.70 x 10^-2)

C = 9.93 x 10^-12 F ......Ans

Charge, Q = C V

Q = 1.09 x 10^-9 C ......Ans

Electric field, E = V/d = (110)/(0.70 x 10^-2)

E = 15714.3 N/C .....Ans

energy Stored, U = C V^2 /2 = (9.93 x 10^-12)(110^2) / 2

= 6 x 10^-8 J .....Ans

battery is disconnected hence charge on plates will not change.

as dielectric material inserted, capacitance increases.

C' = k C = 2.48 x 10^-11 F .......Ans

now V' = Q / C' = 1.09 x 10^-9 / 2.48 x 10^-11

V' = 43.9 Volt

E = V' / d = 6272.5 N/C .........Ans

Energy stored, U = C' V'^2 /2

= (2.48 x 10^-11)(43.9^2) /2

= 2.39 x 10^-8 J .......Ans

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