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 $30m$ from it, would equal.

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

Answer 1

The correct option is C $420N/C$

Equivalent capacitance in the above branch will be $\frac{4(9+3)}{4+9+3}\mu F=3\mu F$.

Total charge in above branch will be $Q=CV=24\mu C$. This charge resides on the $4\mu F$ capacitor and $12\mu F$ (combination of $3\mu F$ and $9\mu F$) capacitor.

Now, voltage across $12\mu F$ combination of capacitors is given by $V=Q/C=24/12=2V$. This is the same as the voltage across $9\mu F$ capacitor.

Hence, charge on $9\mu F$ capacitor is $Q=CV=9\times 2=18\mu C$

From above total charge on $4\mu F$ and $9\mu F$ capacitors is $24+18=42\mu C$

Now, by coulomb's law,

$E=\frac{kQ}{r^2}$

$E=\frac{9\times {10}^9\times 42\times {10}^{-6}}{{30}^2}=420N/C$