An oil drop 'B'has charge 1.6*10*-19 and mass 1.6*10*-14 kg. If the drop is in equilibrium position, then what will be potential difference between the plates. ( distance between the two plates = 100mm)
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Thus, the potential difference between the plates of both capacitors is VA - VB = Vbat. We have C1 = Q1/Vbat and C2 = Q2/Vbat, where Q1 is the charge on capacitor C1, and Q2 is the charge on capacitor C2. Let C be the equivalent capacitance of the two capacitors in parallel, i.e. C = Q/Vbat, where Q = Q1 + Q2.
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