A parallel plate air capacitor has a capacitance C. When it is half filled with a dielectric of dielectric constant 5, the percentage increase in the capacitance will be

Options

A. 4
B. 0.666
C. 0.333
D. 200

Answer

D

Explanation

Initial capacity `C=(epsi_(0)A)/(d)`
When the dielelectric is introduced. We get two capacitors
one of area `(A)/(2) and K=1` and the other of area `(A)/(2)` and K=5,
But d remains constant,
`therefore C'=(epsi_(0)A)/(2d)+(epsi_(0)(5A))/(2d)=(epsi_(0)A)/(2d)(5+1)=(3epsi_(0)A)/(d)`
`therefore Delta C=C'-C=(3epsi_(0)A)/(d)-(epsi_(0)A)/(d)=(2epsi_(0)A)/(d)`
`therefore`The percentage change in C.
`=(DeltaC)/(C)xx100=((2epsi_(0)A)/(d))/((epsi_(0)A)/(d))xx100=200%`.