Explore Westonci.ca, the leading Q&A site where experts provide accurate and helpful answers to all your questions. Discover comprehensive answers to your questions from knowledgeable professionals on our user-friendly platform. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A 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 visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.
