#### Question

A parallel-plate capacitor has plate area 100 cm^{2} 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.

#### Solution

The given system of the capacitor will behave as two capacitors connected in series.

Let the capacitances be *C*_{1} and *C*_{2}.

Now,

`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)`

= `(∈_0A(k_1+k_2))/(k_1d_2+k_2d_1)`

= `((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

Is there an error in this question or solution?

Solution 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) Concept: Capacitors and Capacitance.