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

Question

TestHub · Accepted Answer

C e q = 5 μ F Potential of 4 μF = 6 volt ∴ charge on 4 μF q 4 = 24 μC Potential of 9 μF = 2 volt ∴ charge on 9 μF q 9 = 18 μC Total charge ( q ) = 42 μC E = k q r 2 = 9 × 10 9 × 42 × 10 - 6 900 = 420 N / C

Physics - Capacitor Question with Solution | TestHub

Options:

Doubts & Discussion