Capacitance: How to get 50 microF w/ 80.0 microF

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To achieve a total capacitance of 50 microF using an 80 microF capacitor, the discussion emphasizes using the formula for capacitors in series. The user initially attempted to apply the wrong equations and expressed confusion about the relationship between charge and capacitance. The correct approach involves using the series capacitance formula, C_eq = (C1 * C2) / (C1 + C2), to find the additional capacitor needed. The conversation highlights the importance of understanding the correct formulas for combining capacitors to achieve the desired capacitance. Ultimately, solving the problem correctly requires applying the appropriate series capacitance equation.
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Hey gang,
I am doing a Mastering physics question, and it's got me stumped. ><

Q: You need a capacitance of 50.0 microF, but you don't happen to have a 50.0 microfarads capacitor. You do have a 80.0 microF capacitor.

What additional capacitor is needed to get a capacitance of 50 microF?

I was thinking it has something to do w/ C = Q / deltaV. But that would mean that I would need to have some sort of knowledge of Q. I then tried

Ceq = (1/C1 +1/C2)^-1
and solved for 1/C1 to get 49.9875 microF. But that doesn't seem logical (+ I got it wrong ><)

Any suggestions / help? Greatly appreciated!
 
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Well you obviously solved for the unknown incorrectly. Why not try to solve the simpler equation (assuming series capacitors):

C_{eq} = \frac{C1 \cdot C2}{C1+C2}
 
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