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A parallel plate capacitor is filled by a dielectric whose relative permittivity varies with the applied voltage (U) as \varepsilon =\alpha U where \alpha =2V-1. A similar capacitor with no dielectric is charged to U_{0}=78 V. It is then connected to the uncharged capacitor with the dielectric. Find the final voltage on the capacitors.

Answers (1)

Here, the capacitors are connected in parallel. Therefore, the voltage across the capacitors is the same. Let us assume the final voltage is U. Here, C is the capacitance of the capacitor without the dielectric. At this point, the charge is Q_{1}=CU

If the initial charge is Q_{0} given by Q_{0}=CU_{0}

The conversion of charges is

Q_{1}=Q_{1}+Q_{2}

CU_{0}=CU+\alpha CU_{2}

\alpha U_{2}+U-U_{o}=0

Solving the equation, we get U=6V

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