What Potential Difference Is Needed for Series Capacitors to Store Same Energy?

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To determine the potential difference needed for two capacitors in series to store the same energy as when they are in parallel, the effective capacitance for the series configuration must be calculated using the formula 1/C = 1/C1 + 1/C2. The energy stored in the parallel configuration is given as 0.281 J. The equation for potential difference, ΔV = sqrt(2 * Energy stored / C), can be applied once the correct effective capacitance for the series connection is found. The confusion arises from incorrectly using the parallel capacitance instead of the series capacitance. Correct calculations will yield the required potential difference for the series setup.
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Two capacitors, C1 = 20 µF and C2 = 5.0 µF, are connected in parallel and charged with a 150 V power supply

A.) Energy Stored I found to be is 2.81*10^-1 J

B.)What potential difference would be required across the same two capacitors connected in series in order that the combination store the same energy as in (a)?

Delta V= sqrt of (2*Energy stored/C) This eqn can b used to solve for delta V

I have no idea what I am doing wrong! I have used the capacitor in parallel combo of C= C1+C2... Then I have converted the microFaraday charges of these sums to Faraday but no success

Any Suggestions??
 
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You know that the effective capacitance of capacitors in series is given by:

\frac{1}{C}=\frac{1}{C_1} + \frac{1}{C_2}...

You also know that the energy stored is given by:

E=\frac{1}{2}CV^2

So you have rearranged perfectly well but the problem could just be that you are using the parallel combination rather than the series combination to find the effective capacitance.
 
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