Question

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 opposite 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.

Answer 1

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