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

The Correct Option is C

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