Question

It takes 52,000 Joules to heat a cup of coffee to boiling from room temperature. How long a piece of 20 cm wide Aluminum foil would it take to make a capacitorlarge enoughto hold this amount of energy if one were to use plastic garbage bag with a 2.6 x 10-5m thickness that breaks down at 610 volts as a dielectric

Answer 1

L = 1.11 x
`10^(6)` m, is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.

Step-by-step explanation:

Solution:

Data Given:

Heat Energy = 52000 J

Dielectric Constant of the plastic Bag = 3.7 = K

Thickness = 2.6 x
`10^(5)` m =d

V = 610 volts

A = width x Length

width = 20 cm = 20 x
`10^(-2)` m

Length = ?

So,

we know that,

U = 1/2 C Δ
`v^(2)`

U = 52000 J

C = ?

V = 610 volts'

So,

U = 1/2 C Δ
`v^(2)`

52000 J = (0.5) x (C) x (
`610^(2)`)

C = 0.28 F

And we also know that,

C =
`(K*E*A)/(d)`

E = 8.85 x
`10 ^(-12)`

K = 3.7

A = 0.20 x L

d = 2.6 x
`10^(5)` m

Plugging in the values into the formula, we get:

0.28 =
`(3.7 * 8.85 .10^(-12) * (0.20 . L) )/(2.6 . 10^(5) )`

Solving for L, we get:

L = 1.11 x
`10^(6)` m,

is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.