A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a…

A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due to a point charge $Q$ (having a charge equal to the sum of the charges on the $4 \mu F$ and $9 \mu F$ capacitors), at a point distant $30$ m from it, would equal:
  1. 420 N/C
  2. 480 N/C
  3. 240 N/C
  4. 360 N/C

Solution



$C_{eq} = 5 \mu F$ $Potential\ of\ 4 \mu F = 6 volt$ $\therefore charge\ on\ 4 \mu F\ (q_4) = 24 \mu C$ $Potential\ of\ 9 \mu F = 2 volt$ $\therefore charge\ on\ 9 \mu F\ (q_9) = 18 \mu C$ $Total\ charge\ (q) = 42 \mu C$ $E = $\frac{kq}{r^2}$ = \frac{9 \times 10^9 \times 42 \times 10^{-6}}{900} = 420 N/C$

Asked in: JEE Main 2016 (03 Apr)

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