A parallel-plate capacitor has plate area 100 cm2 and plate separation 1⋅0 cm. A glass plate (dielectric constant 6⋅0) of thickness 6⋅0 mm and an ebonite plate (dielectric constant 4⋅0) are inserted one over the other to fill the space between the plates of the capacitor. Find the new capacitance.
The given system of the capacitor will behave as two capacitors connected in series.
Let the capacitances be C1 and C2.
`C_1 = (∈_0Ak_1)/d_1` and `C_2 = (∈_0Ak_2)/d_2`
Thus, the net capacitance is given by
`C = (C_1C_2)/(C_1+C_2)`
= `((∈_0Ak_1)/d_1 xx (∈_0Ak_2)/d_2)/((∈_0Ak_1)/d_1 +(∈_0Ak_2)/d_2)`
= `((8.85 xx 10^-12) xx (10^-2) xx 24) / ((6 xx 4 xx 10^-3 + 4 xx 6 xx 10^-3)) = 4.425 xx 10^-11 C`
= 44.25 pF