A homemade capacitor is assembled by placing two 10 in pie pans 11 cm apart and

Question

A homemade capacitor is assembled by placing two 10 in pie pans 11 cm apart and connecting them to the opposite terminals of a 9 V battery.

(A) Estimate the capacitance. (pF)

(B) Estimate the charge on each plate. (pC)

(C) Estimate the electric field halfway between the plates. (V/m)

(D) Estimate the work done by the battery to charge the plates. (J)

(E) Which of the above values change if a dielectric is inserted? (Select all that apply.)

a) work done by the battery

b) electric field

c) capacitance

d) charge

Explanation / Answer

(a)
the capacitence is

C = A/d

= 8.854 * 10^-12 ( Pi * (4.5in*0.0254 m/in)^2/ 0.11m

= 3.3 * 10^-12 F

= 3.3 pF

b)

Q = CV = 3.3 * 10^-12 C(9v) = 29.71 pC

c)

E = V/d = 9V/0.11 m=81.81V/m

d)

W = 1/2 * C * V^2 =(1/2)3.3 * 10^-12 F (9V)^2 = 133.65 pJ

e)

if dielectric is inserted capacitence changes thus the charge andwork done

so(a)(c)(d)are change

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