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Q.

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 $4 μF$ and $9 μF$ capacitors), at a point distant 30m from it, would equal

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a

$48 N / C$

b

$240 N / C$

c

$360 N / C$

d

$420 N / C$

answer is D.

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Detailed Solution

$Q = C_{eff} V Q = 5 \times 8 = 40 μ C Change on 4 μF Q_{1} = (\frac{C_{1}}{C_{1} + C_{2}}) Q = (\frac{3}{3 + 5}) (40) = \frac{3}{5} \times 40^{8} \Rightarrow 24 μ C Change on 9 μF Q 3 = (\frac{9}{3 + 9}) 24 = \frac{9}{12} \times 24 \Rightarrow 18 μC Q = 24 + 18 = 42 μ C E = \frac{1}{4 {πε}_{0}} \frac{Q}{r^{2}} = 420 N / C$

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