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Answer :
The potential difference across the 6 μF capacitor is V= 3.71 V.
How to solve
In a series circuit, the total capacitance (C_total) is given by the reciprocal of the sum of the reciprocals of the individual capacitances.
Thus, [tex]C_{total[/tex] = 1/(1/4 + 1/6 + 1/8) μF = 1.714 μF.
Since they are in series, the same charge (Q) will store on each capacitor.
The charge stored on a capacitor is given by Q = CV.
Therefore, Q = 1.714 μF * 13 V = 22.282 μC.
The potential difference (V) across a capacitor is given by V = Q/C.
Therefore, the potential difference across the 6 μF capacitor is V = 22.282 μC / 6 μF = 3.71 V.
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