Two large, parallel conducting plates carrying op­posite charges of equal magnitude are separated by $2.20$ cm. What is the potential difference between the two plates?

Verified step by step guidance

1. Understand that the potential difference between two parallel plates is related to the electric field and the distance between the plates. The formula to use is: $V=E\times d$, where $V$ is the potential difference, $E$ is the electric field, and $d$ is the distance between the plates.
2. Recall that for parallel plates, the electric field $E$ is uniform and can be calculated using the formula: $E=\frac{\sigma }{\epsilon }$, where $\sigma$ is the surface charge density and $\epsilon$ is the permittivity of free space.
3. Determine the surface charge density $\sigma$ if not given, using the relationship between charge $Q$ and area $A$: $\sigma =\frac{Q}{A}$.
4. Substitute the value of $E$ into the potential difference formula: $V=\frac{\sigma }{\epsilon }\times d$.
5. Convert the distance from centimeters to meters by dividing by 100, since SI units are required for calculations: $d=\; 2.20\times 10-2m$. Use this value in the formula to find the potential difference.

Key Concepts

Electric Field Between Parallel Plates

The electric field between two parallel plates is uniform and can be calculated using the formula E = σ/ε₀, where σ is the surface charge density and ε₀ is the permittivity of free space. This uniform field is crucial for determining the potential difference between the plates.

Potential Difference

The potential difference (voltage) between two points in an electric field is the work done to move a unit charge between those points. For parallel plates, it is calculated using V = Ed, where E is the electric field strength and d is the separation distance between the plates.

Capacitor Basics

A capacitor consists of two conductive plates separated by an insulator, storing energy in the electric field between the plates. Understanding capacitors helps in analyzing the relationship between charge, voltage, and electric field, which is essential for solving problems involving parallel plates.