Three capacitors are connected in series to a dc source, as shown in figure. The charges accumulated on the…

Three capacitors are connected in series to a dc source, as shown in figure. The charges accumulated on the plates of the capacitor are \(q_{1}, q_{2}, q_{3}, q_{4}, q_{5}\) and \(q_{6} .\) Then
  1. \(\mathrm{q}_{2}+\mathrm{q}_{4}+\mathrm{q}_{6}=\frac{50}{12} \mathrm{C}\)
  2. \(\mathrm{q}_{1}+\mathrm{q}_{3}+\mathrm{q}_{5}=\frac{50}{12} \mathrm{C}\)
  3. \(\mathrm{q}_{2}+\mathrm{q}_{4}+\mathrm{q}_{6}=0\)
  4. \(\mathrm{q}_{2}=\mathrm{q}_{4}=\mathrm{q}_{6}\)

Solution

When the capacitors are connected in series, same amount of charge (q) gets deposited on the plates of the capacitor.
The charge distribution on the plates is as shown in figure. From the figure it can be seen that the charge developed \(\mathrm{q}_{2} \mathrm{q}_{4} \mathrm{q}_{6}\) on the plates 2,4 and 6 is the same. .

Asked in: JEE Mains - Capacitance - Chapter Test

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