A parallel-plate capacitor is made from two aluminum-foil sheets, each 2.40 cm wide and 13.0 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 that can be stored in this capacitor? (The dielectric constant of mica is 5.4, and its dielectric strength is 1.00 108 V/m.)
I know that C = (e_0 * A)/d
and C = kC_0
So wouldn't it be C = (k*e_0*A)/d? I tried that and I got .663 mC. But it was incorrect. Can someone please help me and explain it? Thank you.
I know that C = (e_0 * A)/d
and C = kC_0
So wouldn't it be C = (k*e_0*A)/d? I tried that and I got .663 mC. But it was incorrect. Can someone please help me and explain it? Thank you.
-
Your formula gives you capacitance in Farads, not charge in Coulombs.
Also be careful with your prefixes - m is for milli (10^-3) , not micro (10^-6).
So the capacitance is 6.627*10^-7 F = 0.663 μF
Now use Q = C*V
The maximum voltage across the capacitor is
V = 1*10^8 V/m * 0.0000225m = 2250V
So the maximum charge is
Q = 6.627*10^-7 F * 2250V = 1.49*10^-3 C = 1.49 mC
Also be careful with your prefixes - m is for milli (10^-3) , not micro (10^-6).
So the capacitance is 6.627*10^-7 F = 0.663 μF
Now use Q = C*V
The maximum voltage across the capacitor is
V = 1*10^8 V/m * 0.0000225m = 2250V
So the maximum charge is
Q = 6.627*10^-7 F * 2250V = 1.49*10^-3 C = 1.49 mC