Animated Solution for Physics - Electrostatics: If there are $n$ capacitors in parallel connected to $V$ volt source, then the energy stored is equal to

Energy Storage in Capacitor Networks

Imagine you are tasked with designing a circuit that needs to store a massive amount of electrical energy, but you only have a handful of standard, low-capacity capacitors at your disposal. How do you maximize your storage? The answer lies in the elegant physics of parallel circuits. Let's dive into a classic problem that explores exactly this scenario.

We are given $n$ identical capacitors, each with a capacitance of $C$. These capacitors are wired in parallel and connected across a battery that provides a steady voltage $V$. Our goal is to determine the total energy stored in this entire network.

The Parallel Advantage

When capacitors are connected in parallel, they all share the same two electrical nodes. This means the potential difference (voltage) across every single capacitor in the network is exactly the same as the source voltage, $V$.

Think of a capacitor as a parking lot for electrons. Connecting them in parallel is like opening up multiple parking lots side-by-side. The total capacity to park cars (store charge) simply adds up. Mathematically, the equivalent capacitance $C_{net}$ of a parallel combination is the algebraic sum of the individual capacitances:

Since we have $n$ identical capacitors, each with capacitance $C$, the math becomes beautifully simple. We just add $C$ to itself $n$ times:

By placing them in parallel, we have effectively created one giant super-capacitor with $n$ times the capacitance of a single unit!

Energy in an Electric Field

Now that we have simplified our complex network into a single equivalent capacitor $C_{net}$, we can calculate the energy it holds. The energy $E$ stored in the electric field of a capacitor is given by the fundamental formula:

This equation tells us that the stored energy scales linearly with the capacitance but quadratically with the voltage.

Bringing It Together

All that remains is to substitute our equivalent capacitance back into the energy equation. We found that $C_{net}=nC$. Plugging this in, we get:

Rearranging slightly for clarity, we arrive at our final, elegant result:

This perfectly matches option (b).

It is fascinating to consider the alternative. What if we had connected these $n$ capacitors in series instead? In a series circuit, the equivalent capacitance drops drastically to $C/n$. The energy stored would then be a mere $\frac{C{V}^2}{2n}$. This stark contrast highlights why parallel configurations are the go-to choice when you need to maximize energy storage in a circuit!