The answer is actually True, I just took the test and it was correct.
C.
Thanks me later, that's my answer.
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
Explanation:
it is energy that flows of electric charge it has the ability to do work or apply force to move an object.
