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 30 m from it , would equal: [JEE Main Solved Paper-2016 ]

A)  480 N/C                              

B) 240 N/C

C) 360 N/C                               

D) 420 N/C

Correct Answer: D

Solution :

\[\text{Q}=\text{24}+\text{18}=\text{42}\mu \text{c}\]                               \[E=\frac{KQ}{{{r}^{2}}}\] \[\Rightarrow \]\[E=\frac{9\times {{10}^{9}}\times 42\times {{10}^{-6}}}{{{(30)}^{2}}}=420N/C\]