Q.Capacity of a parallel capacitor with dielectric constant 5 is 40 $\mu$F. Calculate the capacity of the same capacitor when dielectric material is removed.

Inserting a dielectric slab of thickness t (constant k) into a parallel plate capacitor's gap d changes its capacitance to $C=\dfrac{\epsilon_0A}{(d-t)+t/k}$ -- always greater than the bare $C_0=\epsilon_0A/d$, since $k>1$ shrinks the effective denominator. If the slab fills the ENTIRE gap ($t=d$), this simplifies to $C=kC_0$, the capacitance rising by exactly the factor k -- giving a direct experimental way to measure a material's dielectric constant. …