A combination of capacitors is set up as shown in the figure. The magnitude of the electric field, due of 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 distant 30 m from it, would equal :

Options

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

Answer

B, C, D

Explanation

The equivalent capacitance of 3`mu`F and 9 `mu` F = 12`mu`F Now, the equivalent capacitance of `4mu`F and
`12 mu F = (4xx12)/(4+12)= (48)/(16) = 3 mu F `
`3 mu F ` and `2 mu F ` are in parallel . So P.D. = 8V
`:.` The charge upon `3 mu F q=3 xx10^(-6) xx8 =24 mu C `
`:. ` The charge upon the capacitor of `4 mu F q = 24 mu C `
If the charge upon `12 mu F ` is `24 mu C` then the charge upon `9 mu F , q_(1) =(24 xx9)/(12)= 18 mu C ` and the charge upon `3 mu , q_(2) = 24 -18 = 6 mu F `
`:. Q = 24 + 18 = 42 mu C `
`:. ` Electric field E `= (KQ)/(r^(2))`
`= (9xx10^(9)xx42xx10^(-6))/((30)^(2))`
`= (9xx10^(9)xx42xx10^(-6))/(9xx10^(2))`
`= 42 xx10= 420 (N)/(C)`