https://file.aiquickdraw.com/mathbot/file/17264499918060wygjgxd.jpg

In the problem, a capacitor is created by two metal plates with dimensions \( L = 0.36 \, \text{m} \) and \( W = 0.24 \, \text{m} \). The plates are separated by a distance \( d = 0.1 \, \text{m} \), and are connected to a battery such that the first plate has a charge of \( q \).

### Part (a) of the problem asks:
Write an expression for the magnitude of the electric field, \( E \), halfway between the plates.

### Solution:
For a parallel plate capacitor, the magnitude of the electric field between the plates is uniform and given by:

\[
E = \frac{\sigma}{\epsilon_0}
\]

Where:
- \( \sigma \) is the surface charge density on the plates.
- \( \epsilon_0 \) is the vacuum permittivity constant, \( \epsilon_0 = 8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2 \).

The surface charge density \( \sigma \) is related to the charge \( q \) and the area \( A \) of each plate by:

\[
\sigma = \frac{q}{A}
\]

The area \( A \) of each plate is:

\[
A = L \times W = 0.36 \, \text{m} \times 0.24 \, \text{m} = 0.0864 \, \text{m}^2
\]

Thus, the magnitude of the electric field halfway between the plates is:

\[
E = \frac{q}{A \cdot \epsilon_0}
\]

Substituting the values:

\[
E = \frac{q}{0.0864 \times 8.85 \times 10^{-12}} \, \text{N/C}
\]

This gives the general expression for the electric field \( E \), dependent on the charge \( q \).

---

Would you like further clarification or details on any part of the solution? 

Here are some related questions to consider:

1. What happens to the electric field if the charge on the plate is doubled?
2. How does changing the separation distance \( d \) between the plates affect the electric field?
3. How would the expression for \( E \) change if the plates were placed in a dielectric medium instead of a vacuum?
4. What is the energy stored in the capacitor if the charge \( q \) is known?
5. How does the size of the plates (i.e., \( L \) and \( W \)) affect the capacitance of the system?

Tip: The electric field in a capacitor is directly proportional to the charge and inversely proportional to the plate area!

A capacitor is created by two metal plates. Each plate has dimensions L = 0.36 m and W = 0.24 m. The two plates are separated by a distance, d = 0.1 m, and are connected to a battery such that the first plate has a charge of q. Write an expression for the magnitude of the electric field, E, halfway between the plates.

Learn how to calculate the magnitude of the electric field halfway between two parallel metal plates with known dimensions (L = 0.36 m, W = 0.24 m) and charge q. This physics problem involves key concepts such as capacitance, surface charge density, and Gauss's Law. Ideal for undergraduate physics students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation