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. (a) `420N//C`
B. (b) `480N//C`
C. (c) `240N//C`
D. (d) `360N//C`

Answer

A

Explanation

Charge on `C_1` is `q_1=[((12)/(4+12))xx8]xx4=24muc`
The voltage across `C_P` is `V_P=(4)/(4+12)xx3=2v`
`:.` Voltage across `9muF` is also 2V
`:.` Charge on `9muF` capacitor `=9xx2=18muC`
`:.` Total charge on `4muF` and `9muF=42muc`
`:.E=(KQ)/(r^2)=9xx10^9xx(42xx10^-6)/(30xx30)=420Nc^-1`