Q.
A dielectric of thickness 5 cm and dielectric constant 10, is introduced in between the plates of a parallel plate capacitor having plate area 500 cm2 and separation between plates 10 cm. The capacitance of the, capacitor is (Ifε0=8.8×10−12SIunits):
The capacitance of parallel plate capacitor when dielectric constant introduced between the plates is C=d−t(1−K1)ε0A(Here:K=10,t=5cm,A=500cm2,d=10cm,ε0=8.8×10−12SI unit) Putting the given value in above equation C=10×10−2−5×10−2(1−101)8.8×10−12×500×10−4=10×10−2−4.5×10−128.8×10−12×500×10−4=5.5×10−128.8×10−12×500×10−4=8×10−12F=8pF