A slab of material of dielectric constant K has the same area as the plates of a parallel capacitor, but has a thickness `((3)/(4) d)`,
where d is the separation of the plates. How is the capacitance changed when the slab is inserted between the plates

Explanation

` V = E_o ((d)/(4)) + (E_o)/(K) ((3d)/(4)) = E_o d ((K+3)/(4K)) `
` V = V_o (( K + 3 )/(4K)) `
` C = (Q_o)/(V) = (4K)/(K + 3) (Q_o)/(V_o) = (4 K )/(K + 3 ) Co`