In the given figure the capacitor circuit continues to infinity. The battery connected has e.m.f. equal to V. The potential difference between points 1 and 2 is V2, that between points 3 and 4 is V4 and so on; i.e., the potential difference becomes 12 after every step of the ladder. Find the ratio C1C2.

Let the equivalent capacitance be C0. Even if we remove one ladder, the equivalent capacitance will remain C0.The potential drop across C1 in figure shown is V2.∴ q0C1=V2⇒q0=C1V2                     …(i)Also q0=C0V                                           …(ii)∴ C12=C0⇒C1=2C0                       …(iii)From the given circuit, C0=C0+C2C1C0+C2+C1⇒ C02+C2C0−C1C2=0Which gives, C0=−C2±C22+4C1C22Only positive sign is acceptable.∴ C0=−C2+C22+4C1C22∴ C12+C22=C22+4C1C22 ∵C12=C0C1+C22=C22+4C1C2C12=2C1C2⇒C1=2C2