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

Answer:

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.

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 = $\frac{K*E*A}{d}$

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 = $\frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }$

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.