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 distance 30 m from it, would equal:

Options

A. `240N//C`
B. `360N//C`
C. `420N//C`
D. `480N//C`

Answer

C

Explanation

Resultnt circuit,
As, charge on `3muF=3muF xx8V=24muC therefore `Charge on `4muF=`Charge on `12muF=24muF`.

Charge on `3muF=3muFxx2V=6muC,` Charge on `9muF=9muFxx2V=18muC`
Charge on `4muF+`charge on `9muF=(24+18)muC=42muC`
`therefore`Electric field at a point distant 30m=`(9xx10^(3)xx42xx10^(-6))/(30xx30)=420N//C`.