The capacitance of a parallel plate air capacitor is C. Now, half the areas of its plates are vertically dipped in an oil of dielectric constant 1.6. What would be its capacitance?

Explanation

The upper half would still be an air capacitor. As the plate areas are halved, its capacitance, `C_1 = (C )/(2)`. The lower half, immersed in oil, would have a capacitance, `C_2= (kC)/(2)`. `C_1 " and "C_2` would form a parallel combination, the equivalent capacitance,
`C'= C_1+C_2= (C )/(2)+(kC)/(2)= (C )/(2) (1+k)`
`= (C )/(2)(1+1.6)= 1.3 C`.