Question

A charged capacitor will discharge slowly, even if it's terminal are 'open circuit' that is ,not connected to anything this occurs because the dielectric between the two plates of the capacitor acts as a very poor conductor, allowing a very small leakage current to flow between the two plates,thus discharging them
A capacitor consists of two plates of area 50cm^2 ,separated by a sheet of polythene 0.1mm thick.the capacitor is briefly connected to a power supply, producing a potential difference between the plates of 15v calculate
A.the capacitance of the capacitor
B.the electrical resistance of the
dielectric between the two plates
C.the initial leakage current through the
capacitor when it is disconnected from the power supply

Answer 1

A. The capacitance of the capacitor is given by C = εA/d, where ε is the permittivity of the dielectric, A is the area of the plates, and d is the distance between the plates. In this case, C = 8.85 x 10^-12 F.

B. The electrical resistance of the dielectric between the two plates is given by R = ρL/A, where ρ is the resistivity of the dielectric, L is the length of the dielectric, and A is the area of the plates. In this case, R = 1.78 x 10^14 Ω.

C. The initial leakage current through the capacitor when it is disconnected from the power supply is given by I = V/R, where V is the potential difference between the plates and R is the electrical resistance of the dielectric. In this case, I = 8.45 x 10^-15 A.