A parallel plate capacitor has square plates with sides of length 20 cm. The dis

Question

A parallel plate capacitor has square plates with sides of length 20 cm. The distance between the plates is 4 mm. The plates are charged up to 50volts.

a.What is the electric field between the plates? (N/C)

b.What is the amount of charge on each plate? (C)

c. What is the capacitance? (?F)

d.What is the energy stored by the capacitor? (nJ)

eThe space between the plates is filled with a dielectric which has a dielectric constant of 20, and the new capacitor is charged up to the same voltage as before. How much energy is stored in the capacitor now?(nJ)

thanks

Explanation / Answer

area of plates = 0.2*0.2 = 0.04 m^2

distance d = 0.004 m

V = 50 volts

a) E = V/d = 50 / 0.004 = 12500 N/m

b)Q = CV

C = A*E0 / d (E0 is epsilon not )

C = 8.85*10^-11 C

so

Q = 4.425 nC

c)C = 8.85*10^-11 C

d)Energy = 1/2 QV = 1.106*10^-7 J

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