A capacitor is formed by two square metal-plates of edge a, separated by a distance d. Dielectrics of dielectric constant K_{1} and K_{2}_{ }are filled in the gap as shown in figure . Find the capacitance.

#### Solution

Let us consider an elemental capacitor of width dx at a distance x from the left end of the capacitor. It has two capacitive elements of dielectric constants K_{1} and K_{2}_{ }with plate separations (x tan θ) and (d − x tan θ) in series, respectively. The areas of the plates of the capacitors are adx.

The capacitances of the capacitive elements of the elemental capacitor are :

`dC_1 = (∈_0K_2(adx))/(x tan θ) , dC_2 = (∈_0K_1(adx))/(d - x tan θ)`

The net capacitance of the elemental capacitor is given by

`1/(dC)= 1/(dC_1) + 1/(dC_2)`

`1/(dC) = (x tanθ)/(∈_0K_2(adx)) + (d-x tanθ)/(∈_0K_1(adx))`

`⇒ dC = (∈_0K_1K_2(adx))/(K_1 x tanθ + K_2(d-x tanθ)`

Thus, integrating the above expression to calculate the net capacitance

`C = ∫_0^a dC = ∫_0^a (∈_0K_1K_2adx)/(K_1x tanθ+ K_2(d-x tanθ)`

`⇒ C = ∈_0K_1K_2a ∫_0^a (dx)/(K_2d + x tanθ(K_1 - K_2)`

`⇒ C = (∈_0K_1K_2a)/(tanθ(K_1 - K_2))[log_e[K_2d + x tanθ(K_1 - K_2) ]]_0^a`

`⇒ C = (∈_0K_1K_2a)/(tan θ(K_1 - K_2)) [ log_e[K_2d + a tan θ(K_1 - K_2)] - log_e K_2d]`

As we know that `tan θ = d/a` substituting in the expression for capacitance *C*.

Now,

`⇒ C = (∈_0K_1K_2a)/(d/a xx (K_1 - K_2) )[ log_e[K_2d + a xx d/a(K_1 - K_2)] - log_e K_2d]`

`⇒ C = (∈_0K_1K_2a)/(d/a xx (K_1 - K_2) )[ log_e K_1d - log_e K_2d ]`

`⇒ C = (∈_0K_1K_2a^2)/(d(K_1 - K_2)) [log_e (K_1/K_2)]`