A parallel plate capacitor separated by an air gap is made of \(0.8m^2\) tin plates and 20 mm apart. It is connected to 120 V battery. What is the charge on each plate?
Take \(ε_o\) = \(8.85×10^-12 Fm^-1\)
A.
3.54nC
B.
42.5nC
C.
35.4nC
D.
4.25nC
Correct answer is B
A= \(0.8m^2\) d= 20mm =\(\frac{20}{1000}\) = 0.02m
v =120v; \(ε_oA\)= \(8.85 × 10^-12 fm^-1\)
C = \(\frac{ε_oA}{d}\)
C =\(\frac{8.85 × 10^-12×0.8}{0.02}\)
C= \(3.54 × 10^-10 F\)
Q= CV
⇒ \(3.54 × 10^-10\) × 120 = \(4.25×10^-8c\)
Q=\(42.5 × 10^-9c\) = 42.5nC
