Initial capacitance of the air capacitor is given by:
C=dϵ0A
When the capacitor is half-filled with a dielectric of thickness d/2, it can be considered as two capacitors connected in series.
Capacitance of the dielectric-filled part:
C1=d/2Kϵ0A=d2Kϵ0A=2KC
Capacitance of the air-filled part:
C2=d/2ϵ0A=d2ϵ0A=2C
The equivalent capacitance Ceq of the series combination is:
Ceq=C1+C2C1C2=2KC+2C(2KC)(2C)=K+12KC
Substituting K=5:
Ceq=5+12(5)C=610C=35C
The increase in capacitance is:
ΔC=Ceq−C=35C−C=32C
The percentage increase in capacitance is:
% increase=CΔC×100=32×100≈66.67%
Answer: 66.67