JAMB - Physics (2023 - No. 20)
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 \(ε_o\) = \(8.85×10^-12 Fm^-1\)
Take \(ε_o\) = \(8.85×10^-12 Fm^-1\)
3.54nC
42.5nC
35.4nC
4.25nC
Explanation
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
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
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