Animated Solution for Physics - Electrostatics: 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 $4\mu \text{F}$ and $9\mu \text{F}$ capacitors), at a point distance $30\text{ m}$ from it, would equal to

When faced with a complex circuit, the best approach is to break it down into simpler, manageable chunks. Let's look closely at the structure of this circuit. We have an $8\text{V}$ battery powering the entire system. The $2\mu \text{F}$ capacitor is connected directly across the battery, meaning it operates independently of the top branch. Our primary focus is the top branch, which contains the $4\mu \text{F}$, $3\mu \text{F}$, and $9\mu \text{F}$ capacitors.

Notice the right side of this top branch: the $3\mu \text{F}$ and $9\mu \text{F}$ capacitors are connected in parallel. In a parallel combination, we simply add their capacitances together:

Now, we can visualize this entire parallel section as a single $12\mu \text{F}$ capacitor. This equivalent capacitor is in series with the $4\mu \text{F}$ capacitor. To find the equivalent capacitance of this entire top branch, we use the product-over-sum rule for series capacitors:

This entire top branch, which behaves like a single $3\mu \text{F}$ capacitor, is connected directly across the $8\text{V}$ battery. Using the fundamental capacitor equation $Q=CV$, we can find the total charge drawn by this branch:

In a series circuit, the charge remains constant across all components. This means the $24\mu \text{C}$ of charge must flow directly through the $4\mu \text{F}$ capacitor. Therefore, we have our first crucial piece of information:

To find the charge on the $9\mu \text{F}$ capacitor, we first need to determine the voltage across the parallel section. We know the total charge entering this section is $24\mu \text{C}$, and its equivalent capacitance is $12\mu \text{F}$. The voltage drop across it is:

Since the $3\mu \text{F}$ and $9\mu \text{F}$ capacitors are in parallel, they both experience this $2\text{V}$ potential difference. We can now easily calculate the charge on the $9\mu \text{F}$ capacitor:

The problem asks for the electric field due to a point charge $Q$, which is defined as the sum of the charges on the $4\mu \text{F}$ and $9\mu \text{F}$ capacitors. Let's add them up:

Finally, we calculate the electric field at a distance of $r=30\text{ m}$ using the standard formula for the electric field of a point charge:

Substituting our values ($k=9\times 10^9\text{ N}\cdot {\text{m}}^2/{\text{C}}^2$):

This elegant simplification leads us directly to our final answer.