question_answer

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 30m from it, would equal:

A) 420 N/C

B)        480 N/C

C) 240 N/C

D)        360 N/C

Correct Answer: A

Solution :

[a] Charge on \[{{C}_{1}}\] is \[{{q}_{1}}=\left[ \left( \frac{12}{4+12} \right)\times 8 \right]\times 4\] \[=24\mu C\] The voltage across \[{{C}_{_{P}}}\] is \[{{V}_{P}}=\frac{4}{4+12}\times 8=2V\] \[\therefore \] Voltage across \[9\mu F\] is also 2V \[\therefore \] Charge on \[9\mu F\] capacitor \[=9\times 2=18\mu C\] \[\therefore \] Total charge on \[4\mu F\] and \[9\mu F=42\mu C\] \[\therefore \]\[E=\frac{KQ}{{{r}^{2}}}=9\times {{10}^{9}}\times \frac{42\times {{10}^{-6}}}{30\times 30}=420N{{C}^{-1}}\]