At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
Sagot :
Answer:
L = 1.11 x [tex]10^{6}[/tex] 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 [tex]10^{5}[/tex] m =d
V = 610 volts
A = width x Length
width = 20 cm = 20 x [tex]10^{-2}[/tex] m
Length = ?
So,
we know that,
U = 1/2 C Δ[tex]v^{2}[/tex]
U = 52000 J
C = ?
V = 610 volts'
So,
U = 1/2 C Δ[tex]v^{2}[/tex]
52000 J = (0.5) x (C) x ([tex]610^{2}[/tex])
C = 0.28 F
And we also know that,
C = [tex]\frac{K*E*A}{d}[/tex]
E = 8.85 x [tex]10 ^{-12}[/tex]
K = 3.7
A = 0.20 x L
d = 2.6 x [tex]10^{5}[/tex] m
Plugging in the values into the formula, we get:
0.28 = [tex]\frac{3.7 * 8.85 .10^{-12} * (0.20 . L) }{2.6 . 10^{5} }[/tex]
Solving for L, we get:
L = 1.11 x [tex]10^{6}[/tex] m,
is the length of piece of 20 cm wide Aluminum foil to make capacitor large enough to hold 52000 J of energy.
Thank you for choosing our service. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.
