A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as ε = αU where α = 2V-1. A similar capacitor with no dielectric is charged to U0 = 78 V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.
On connecting the two given capacitors, let the final voltage be V. If capacity of capacitor without the dielectric is C, then the charge on this capacitor is q1=CV The other capacitor with dielectric has capacity εC. Therefore, charge on it is q2=εCV As ε=αV, therefore q2=ε=αV The initial charge on the capacitor (Without dielectric) that was charged is q0=CV0 From the conservation of charge, q0=q1+q2 CV0=CV+αCV2orV2+V−V0=0 V=2α−1±1+4αV0 using α=2V−1 or V0=78V, we get V=2×2−1±1+(4×2×78)=4−1±625 As V is positive, therefore, V=4625−1=424=6V,
