A parallel plate capacitor is filled by a die | vedclass

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(C) Let $C_0$ be the capacitance of the capacitor without the dielectric.The charge on the first capacitor is $Q_0 = C_0 U_0 = 78 C_0$.When connected

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(C) Let $C_0$ be the capacitance of the capacitor without the dielectric.The charge on the first capacitor is $Q_0 = C_0 U_0 = 78 C_0$.When connected in parallel,the final voltage $U$ across both capacitors is the same.The capacitance of the second capacitor with dielectric is $C = \varepsilon_r C_0 = \varepsilon C_0 = (\alpha U) C_0$.The total charge is conserved: $Q_0 = C_0 U + C U = C_0 U + (\alpha U) C_0 U = C_0 U (1 + \alpha U)$.Substituting the values: $78 C_0 = C_0 U (1 + 2U)$.$78 = U + 2U^2 \implies 2U^2 + U - 78 = 0$.Solving the quadratic equation using the formula $U = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$U = \frac{-1 \pm \sqrt{1^2 - 4(2)(-78)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 624}}{4} = \frac{-1 \pm \sqrt{625}}{4} = \frac{-1 \pm 25}{4}$.Since voltage must be positive,$U = \frac{24}{4} = 6 \ V$ is not among options,let's recheck the calculation.Wait,$78 = U + 2U^2$. If $U = 6$,$6 + 2(36) = 78$. The calculation is correct. Given the options,let's re-evaluate the question parameters. If $U_0 = 78$,$U=6$. If the options are $39, 26, 13, 52$,perhaps $\alpha$ is different or $U_0$ is different. Assuming the logic holds,the result is $6 \ V$.

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