Discussion Forum

1)

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 charge on the 4 µF and 9 µF capacitors ) at a point distant 30m from it, would equal to

 

Answer:

Option C

Explanation:

3 μ and 9 μ =12 μF

4μF and 12 μ F = $\frac{4\times 12}{4 \times 12}=3\mu F$

Q =CV =3× 8 =24 μC(on 4 μF and 3 μF )

Now, this 24 μC distributes in direct ratio of capacity between 3μ F and 9 μ F

  Therefore ,Q_9μF =18μC

   $Q_{4\mu F}+Q_{9\mu F}=24+18$

= 42μ C =Q            (say)

$E=\frac{kQ}{R^{2}}=\frac{9\times 10^{9}\times42\times 10^{-6}}{30^{2}}=420 N/C$