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 4μF and 9μF capacitors), at a point distant 30 m from it, would equal :

Engineering

Physics

Parallel Plate Capacitor

Question

420 N/C

480 N/C

360 N/C

240 N/C

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Q = Q₄ + Q₉

= 24 µ C + 18 µ C = 42 µC

$E = \frac{kQ}{r^{2}} = 420 N / C$