Animated Solution for Physics - Electrostatics: Two identical metal plates are given positive charges $Q_1$ and $Q_2$ ($<Q_1$) respectively. If they are now brought close together to form a parallel plate capacitor with capacitance $C$, the potential difference between them is

Have you ever wondered what happens when you take two identical metal plates, give them different amounts of charge, and then bring them close together? It's a classic scenario that tests your understanding of electric fields and capacitance. Let's dive into the physics behind it!

We give the first plate a positive charge $Q_1$ and the second plate a positive charge $Q_2$. We are given a crucial piece of information: $Q_1$ is greater than $Q_2$.

When we bring these plates close together, separated by a small distance $d$, they form a parallel plate capacitor. But wait, a standard capacitor usually has equal and opposite charges ($+Q$ and $-Q$). Here, both are positive and unequal! How do we find the potential difference between them?

Superposition of Electric Fields To find the potential difference, we first need to determine the electric field in the space between the plates. The beauty of electromagnetism is the principle of superposition. The net electric field is simply the vector sum of the fields produced by each plate individually.

Recall that the electric field produced by a single, large, uniformly charged plate is given by: where $A$ is the area of the plate and ${\epsilon }_0$ is the permittivity of free space. This field points away from a positively charged plate.

In the region between our two plates: - Plate 1 creates an electric field $E_1=\frac{Q_1}{2A{\epsilon }_0}$ pointing to the right (towards Plate 2). - Plate 2 creates an electric field $E_2=\frac{Q_2}{2A{\epsilon }_0}$ pointing to the left (towards Plate 1).

Since $Q_1>Q_2$, the field $E_1$ is stronger. The net electric field $E_{net}$ will point to the right and its magnitude will be the difference between the two:

From Field to Potential Now that we have the uniform net electric field between the plates, finding the potential difference $V$ is straightforward. The relationship is:

Substituting our expression for $E_{net}$:

Let's rearrange this equation to reveal a familiar term. We can move $d$ to the denominator's denominator:

Does the term $\frac{A{\epsilon }_0}{d}$ ring a bell? Yes, it is the standard formula for the capacitance $C$ of a parallel plate capacitor!

Substituting $C$ into our equation, we arrive at the elegant final result:

And there you have it! The potential difference depends on the difference between the charges. If the charges were equal ($Q_1=Q_2$), the potential difference would be zero, which makes perfect sense.