Question 62 of 180

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 distant 30 m from it, would equal:

A420 N/CB480 N/CC240 N/CD360 N/C

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Answer: (A) 420 N/C

$Ceq=5\mu F$

$Potentialof4\mu F=6volt$

$∴charge\text{on}4\mu F\left({\text{q}}_{\text{4}}\right)=24\mu C$

Potential of$9\mu F=2volt$
$∴charge\text{on}9\mu F\left({\text{q}}_{9}\right)=18\mu C$
$Totalcharge(q)=42\mu C$
$E=\dfrac{kq}{{r}^{2}}=\dfrac{9\times {10}^{9}\times 42\times {10}^{-6}}{900}=420N/C$