Question

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 :

• A. 240 N/C
• B. 360 N/C
• C. 420 N/C
• D. 480 N/C

Answer 1

Answer: (C)

420 N/C

Answer 2

The equivalent capacitance of the above branch will be $\frac{4(9+3)}{4+9+3}=3\mu F$
The total charge in the above branch will be Q=CV=24 $\mu F$
Now voltage across 12$\mu F$ V= gives the combination of capacitors$\frac{Q}{C}=\frac{24}{12}=2\,V$
This is the same as the voltage across 9$\mu F$ capacitor.
Hence, the charge on 9 $\mu F$ capacitor is =24+18=42$\mu C$
Now by Coulomb's law,
$E=\frac{kQ}{r^2}$
$E=\frac{9\times10^9\times42\times10^{-6}}{30^2}$=420 N/C