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

Answer

C

Explanation

In Fig, `3 muF` and `9 muF` capacitors are in parallel.
`:. C_(p) = 3+9 = 12 muF`
This combination is in series with `4 muF` capacitor
`(1)/(C_s) = (1)/(12) + (1)/(4) = (1)/(3), C_(s) = 3 muF`
`Q = CV = 3xx8 = 24 muC`
This is charge distributies iteself in the direct ratio of capacitros `3 muF` and `9muF`.
`:. Q_(9 muF) = ((24)xx9)/(3+9) = 18 muC`
Now `Q_(9 muF) + Q_(4 muF) = 18+24 = 42 muC`
Electric field intensity due to `Q = 42 muC` at a distance of `30m` from it
`E = (kQ)/(r^(2)) = (9xx10^(9)xx42xx10^(-6))/(30xx30)`
`= 420 N//C`