A parallel-plate capacitor is made from two aluminum-foil sheets, each 2.40 cm wide and 13.0 m longBetween the sheets is a mica strip of the same width and length that is 0.0225 mm thickWhat 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)/dand C kC_0So wouldn't it be C (ke_0A)/d? I tried that and I got .663 mCBut it was incorrectCan someone please help me and explain it? Thank you.
Your formula gives you capacitance in Farads, not charge in CoulombsAlso be careful with your prefixes - m is for milli (10^-3) , not micro (10^-6)So the capacitance is 6.62710^-7 F 0.663 μF Now use Q CV The maximum voltage across the capacitor is V 110^8 V/m 0.0000225m 2250V So the maximum charge is Q 6.62710^-7 F 2250V 1.4910^-3 C 1.49 mC
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