Animated Solution for Physics - Electrostatics: A parallel plate capacitor with plates of area $1{\text{ m}}^2$ each, are at a separation of $0.1\text{ m}$. If the electric field between the plates is $100\text{ N/C}$, the magnitude of charge on each plate is $\left(\text{Take, }{\epsilon }_0=8.85\times 10^{-12}\frac{{\text{C}}^2}{{\text{N-m}}^2}\right)$

The magic of capacitors lies in their ability to store electrical energy in the form of an electric field. In this thrilling problem, we are tasked with finding the magnitude of the charge on the plates of a parallel plate capacitor.

We are given some specific parameters: the area of the plates, the separation distance between them, and the uniform electric field existing in that space. At first glance, it might seem like a straightforward plug-and-chug exercise, but as we will discover, there is a beautiful mathematical elegance hidden within the equations that makes one of our given values completely redundant!

Let us visualize the physical reality of the problem. We have two parallel conducting plates.

The area of each plate is given as $A=1{\text{ m}}^2$.

The separation between the plates is $d=0.1\text{ m}$.

The electric field between the plates is uniform and has a magnitude of $E=100\text{ N/C}$.

Our goal is to find the magnitude of the charge, $Q$, on each plate.

The fundamental relationship that governs any capacitor is the charge equation. The charge $Q$ stored on a capacitor is directly proportional to the potential difference $V$ across its plates, with the constant of proportionality being the capacitance $C$.

This is our master equation. To find $Q$, we need to determine both $C$ and $V$ for our specific parallel plate setup.

For a parallel plate capacitor with a vacuum (or air) between its plates, the capacitance is determined purely by its geometry. It is directly proportional to the plate area $A$ and inversely proportional to the separation distance $d$.

Here, ${\epsilon }_0$ is the permittivity of free space, a fundamental constant of nature.

Next, we need to find the potential difference $V$. We know that for a uniform electric field $E$, the potential difference across a distance $d$ is simply the product of the field strength and the distance.

Now comes the most satisfying part of the physics problem—the algebraic substitution. Let us substitute our expressions for $C$ and $V$ back into the master equation.

Look closely at what happens here. The separation distance $d$ appears in the denominator of the capacitance and in the numerator of the voltage.

The $d$ completely cancels out! This is a profound physical insight. It tells us that if we know the electric field and the area of the plates, the charge is completely independent of how far apart the plates are. The examiner gave us $d=0.1\text{ m}$ as a distractor—a classic trap to test if students blindly plug in numbers or if they simplify their algebra first.

With our simplified, elegant equation, we can now substitute the numerical values.

Multiplying these values is straightforward. The $100$ is simply $10^2$, which when multiplied by $10^{-12}$ gives $10^{-10}$.

This matches option (b) perfectly.

A Gauss's Law Perspective

Could we have solved this even faster? Absolutely! Let us look at the problem through the lens of Gauss's Law.

We know that the electric field between two oppositely charged parallel plates is given by:

Where $\sigma$ is the surface charge density. Since surface charge density is simply the total charge divided by the area ($\sigma =\frac{Q}{A}$), we can rewrite the electric field as:

Rearranging this equation to solve for $Q$ yields:

This is the exact same equation we derived earlier, but we arrived at it in just one step! This highlights the power of understanding multiple physical principles. Whether you use the circuit perspective ($Q=CV$) or the field perspective (Gauss's Law), the physics remains beautifully consistent.