Question

Please no referrals.

A homemade capacitor is assembled by placing two 8-in. pie pans 5 cm apart and connecting them to the opposite terminals of a 9-V battery.

A) Estimate the capacitance.

B) Estimate the charge on each plate.

C) Estimate the electric field halfway between the plates.

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

E) Which of the above values change if a dielectric is inserted?

 

Answer 1

The capcaitance of the capacitor is, EoA 0.0254 m 8.854x102 5x10 -5.74x10-1 F Charge on each plate is, -(5.74x10-12) (9) - 5.16x10-11 c (574x102) 9 The electric field between the plates is 9 V 5x10 -180 V/mm

The work done by the battery is equal to the energy stored in the capacitor is, CV (5.74x10-2)9) 232.5x10- J The capacitance, charge and work done by the battery

Answer 2

Solution)

a)

Here, Area, A = Pi * (8*0.0254)^2 / 4

= 0.0324 m^2

Now,

Capacitance, C= Aεo/d

= 0.0324 * 8.854 * 10^-12 / 0.05 m

= 5.73 * 10^-12 F = 5.73 pF

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b)

We know,

Q = CV = 5.73 * 10^-12 C *9 V= 5.16*10^-11 C

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c)

E = Q/Aεo = 180 N/m

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d)

Work done, W = 1/2 * C * V^2 = 2.32* 10^-10 J

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