Question

The radio tuning capacitor consists of 26 plates of area 24 cm². The distance between plates when set for maximum capacitance is 2mm. Assuming that the area of projection at maximum capacity is equal to the area of the plates, find the maximum capacity of this condenser.

Answer 1

The capacitance is given as follows (see https://www.electronics-tutorial.net/electronic-components/capacitors/multiple-plate-capacitor/):

$$C = \dfrac{\varepsilon\varepsilon_0 (N-1)A}{d}$$

where $\varepsilon=1$ (assuming air between plates), $\varepsilon_0=8.85\times 10^{-12}F/m, A = 24cm^2 = 2.4\times 10^{-3}m^2, N = 26,d = 2mm=2\times10^{-3}m$. Thus, obtain:

$$C = \dfrac{8.85\times 10^{-12}\cdot 25\cdot 2.4\times 10^{-3}}{2\times 10^{-3}} \approx 2.7\times 10^{-10}F$$

Answer. $2.7\times 10^{-10}F$.