This question is incomplete, the complete question is;
A parallel-plate capacitor is made from two aluminum-foil sheets, each 3.0 cm wide and 5.00 m long. Between the sheets is a mica strip of the same width and length that is 0.0225 mm thick. What is the maximum charge?
(The dielectric constant of mica is 5.4, and its dielectric strength is 1.00×10⁸ V/m)
Answer: the maximum charge q is 716.85 μF
Explanation:
Given data;
with = 3.0 cm = 0.03
breathe = 5.0 m
Area = 0.03 × 5 = 0.15 m²
dielectric strength E = 1.00 × 10⁸
∈₀ = 8.85 × 10⁻¹²
constant K = 5.4
maximum charge = ?
the capacitor C = KA∈₀ / d
q = cv so c = q/v
now
q/v = KA∈₀ / d
q = vKA∈₀/d = EKA∈₀
we substitute
q = (1.00 × 10⁸) × 5.4 × 0.15 × 8.85 × 10⁻¹²
q = 716.85 × 10⁻⁶ F
q = 716.85 μF
the maximum charge q is 716.85 μF
Answer:
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B) I just took the test and put D but it gave me the wrong answer. It told me it was B.
Answer:
You need to use charts or a calculator to get 56%
Explanation:
Answer: A
Explanation:
honestly, it sounded the best