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 - I 3-4-2016] |
A) 480 N/C
B) 240 N/C
C) 360 N/C
D) 420 N/C
Correct Answer: D
Solution :
[d] |
\[\text{Q}=\text{24}+\text{18}=\text{42}\mu \text{c}\] \[E=\frac{KQ}{{{r}^{2}}}\] |
\[\Rightarrow \]\[E=\frac{9\times {{10}^{9}}\times 42\times {{10}^{-6}}}{{{(30)}^{2}}}=420N/C\] |
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