Electrostatics 5 Question 18
20. 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 to
(2016 Main)
(a) $240 N / C$
(b) $360 N / C$
(c) $420 N / C$
(d) $480 N / C$
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Solution:
- $3 \mu F$ and $9 \mu F=12 \mu F$
$$ \begin{aligned} 4 \mu F \text { and } 12 \mu F & =\frac{4 \times 12}{4+12}=3 \mu F \\ Q & =C V=3 \times 8=24 \mu C(\text { on } 4 \mu F \text { and } 3 \mu F) \end{aligned} $$
Now, this $24 \mu Cdistributes$ in direct ratio of capacity between $3 \mu F$ and $9 \mu F$. Therefore,
$$ \begin{aligned} Q _{9 \mu F} & =18 \mu C \\ \therefore Q _{4 \mu F}+Q _{9 \mu F}=24+18 & =42 \mu C=Q \\ E & =\frac{k Q}{R^{2}}=\frac{9 \times 10^{9} \times 42 \times 10^{-6}}{30^{2}} \\ & =420 N / C \end{aligned} $$
