What is the capacitance of a parallel plate capacitor with metal plates, each of area 1.50 m2, separated by 2.01 mm?

Question

1. What is the capacitance of a parallel plate capacitor with metal plates, each of area 1.50 m2, separated by 2.01 mm? What charge is stored in this capacitor if a voltage of 2.50 × 103 V is applied to it?

2. The capacitors have values C1 = 2.5 µF and C2 = 4.0 µF, C3 = 6.0 µF C4 = 8.0 µF and the potential difference across the battery is 12.0 V. Assume that the capacitors are connected in series. Find the equivalent capacitance of the circuit and solve for the potential difference across each capacitors.

Answer 1

1.   capacitance of a parallel plate capacitor is C=ε₀A/d   where ε₀=8.85∙10-12 C²/Nm²  A is plate area in m², d plates separation in m.

 Then C=8.85∙10^-12 C²/Nm² )(1.5 m²)/(2.01∙10^-3 m )= 5.5 nF.

Q=CV=(5.5∙10^-9 F)(2.5∙10³ V)=13.7 µF.

2.    C_eq=C1+C2+C3+C4=20.5 µF ; Capacitors in parallel have the same voltage across

each one.