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

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The discussion focuses on determining the potential difference required for two capacitors, C1 = 20 µF and C2 = 5.0 µF, connected in series to store the same energy as when they are connected in parallel with a 150 V power supply. The energy stored in the parallel configuration is calculated to be 0.281 J. The formula for potential difference across capacitors in series, ΔV = √(2 * Energy stored / C), is highlighted as essential for solving the problem. The effective capacitance for capacitors in series is given by the formula 1/C = 1/C1 + 1/C2.

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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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