In a simple electric circuit, a 110-volt electric heater
1 answer:
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Answer:
A) C = 4.425 10⁻¹³ F, B) Q = 6.31 10⁻¹¹ C, C) E = 8.9 10⁴ N / C,
D) u_{E} = 3.19 10⁻⁹ J
Explanation:
Part A. Capacitance is
C = ε₀ A / d
in this case distance is d = 0.800 cm = 0.800 10⁻² m and the area of the plates is x = 0.200 cm= 0.2 10⁻² m, all the quantities must be in the SI system for the result to be in Farads
A = x²
A = (0.2 10⁻²)²
A = 4 10⁻⁶ m
let's calculate
C = 8.85 10⁻¹² 4 10⁻⁶ / 0.8 10⁻²
C = 44.25 10⁻¹⁴ F
C = 4.425 10⁻¹³ F
Part B. The charge on each plate is
Q = C ΔV
Q = 4.425 10⁻¹³ 120
Q = 6.31 10⁻¹¹ C
Part C. the electric field of a plate is
E = σ / 2ε₀
where the charge density is
σ = Q / A
we substitute
E =
let's calculate
E =
E = 8.9 10⁴ N / C
Part D. stored energy
= ½ C V²
u_{E} = ½ 4.425 10⁻¹³ 120²
u_{E} = 3.19 10⁻⁹ J