The figure shows the charge (q) versus voltag | vedclass

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(A) From the relation $q = CV$, the slope of the $q-V$ graph represents the capacitance $C = q/V$.For line $A$, the capacitance is $C_A = 500\,\mu C /

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$40\,\mu F$ and $10\, \mu F$

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(A) From the relation $q = CV$, the slope of the $q-V$ graph represents the capacitance $C = q/V$.For line $A$, the capacitance is $C_A = 500\,\mu C / 10\,V = 50\,\mu F$.For line $B$, the capacitance is $C_B = 80\,\mu C / 10\,V = 8\,\mu F$.Since the parallel combination has a higher equivalent capacitance than the series combination, we have $C_{parallel} = 50\,\mu F$ and $C_{series} = 8\,\mu F$.Let the two capacitors be $C_1$ and $C_2$. Then $C_1 + C_2 = 50$ and $(C_1 C_2) / (C_1 + C_2) = 8$.Substituting $C_1 + C_2 = 50$ into the series formula: $(C_1 C_2) / 50 = 8$, so $C_1 C_2 = 400$.Solving the quadratic equation $x^2 - 50x + 400 = 0$, we get $(x - 40)(x - 10) = 0$.Thus, the capacitances are $40\,\mu F$ and $10\,\mu F$.

The figure shows the charge $(q)$ versus voltage $(V)$ graph for series and parallel combinations of two given capacitors. The capacitances are:

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