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3 years ago

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A homemade capacitor is assembled by placing two 10-in. pie pans 2 cm apart by a piece of glass and connecting them to the oppos

ite terminals of a 12-V battery. Estimate (a) the capacitance, (b) the charge on each plate, (c) the electric field halfway between the plates, and (d) the work done by the battery to charge the plates.

1 answer:

valentina_108 [34]3 years ago

4 0

Answer:

capacitance = 2.242 × $10^{-11}$ F

charge = 1.345 × $10^{-10}$ C

electric field = 600 V/m

work done = 8.07 × $10^{-10}$ J

Explanation:

given data

battery c = 12 V

diameter = 10 in

distance d = 2 cm = 2× $10^{-2}$ m

to find out

capacitance , charge on plate, electric field and work done

solution

we know radius is diameter / 2

so radius r = 10 / 2 = 5 in = 0.127 m

and capacitance formula that is

capacitance = A∈/d

put here all value

capacitance = πr² ∈/d

capacitance = π(0.127)² ×8.85 × $10^{-12}$ /2× $10^{-2}$

capacitance = 2.242 × $10^{-11}$ F

and

charge on plate is express as

charge = capacitance × c

we know here 2 plate so on 1 plate c is 6

charge = 2.242 × $10^{-11}$ × 6

charge = 1.345 × $10^{-10}$ C

and

electric field is express as

electric field = c / d

electric field = 12 / 2× $10^{-2}$

electric field = 600 V/m

and

work done is express as

work done = 1/2 × charge × C

put here value

work done = 1/2 × (1.345 × $10^{-10}$ ) (12)

work done = 8.07 × $10^{-10}$ J

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