In the figure two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge densities of opposite signs and magnitude 6.51 x 10-22 C/m². What is the magnitude of the electric field at points (a) to the left of the plates, (b) to the right of them, and (c) between them?

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### Electric Field between Parallel Metal Plates

In the figure, two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge densities of opposite signs and magnitude \(6.51 \times 10^{-22} \, \text{C/m}^2\).

The problem is to find the magnitude of the electric field at points:
- (a) To the left of the plates
- (b) To the right of them
- (c) Between them

#### Explanation of the Diagram

The diagram shows two parallel plates that are thin and large relative to the distance separating them. The plates are aligned vertically, and the \( x \)-axis runs horizontally through the center of each plate.

1. **Left Plate (Positive Charge)**:
    - The left plate has lines indicating a positive surface charge density.
2. **Right Plate (Negative Charge)**:
    - The right plate has lines indicating a negative surface charge density.

#### Points to Identify

- **(a) To the left of the plates**:
    - This point is located outside and to the left of the positively charged plate.
- **(b) To the right of the plates**:
    - This point is located outside and to the right of the negatively charged plate.
- **(c) Between the plates**:
    - This point is located in the region directly between the positive and negative plates.

### Calculation of Electric Fields

Using the principle that the electric field \( E \) due to a large, flat, uniformly charged plate with charge density \( \sigma \) is given by:

\[ E = \frac{\sigma}{2\epsilon_0} \]

Where:
- \( \sigma \) is the surface charge density
- \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 \approx 8.854 \times 10^{-12} \, \text{Fm}^{-1} \))

Here are the specific calculations for each region:
- **(a) To the left of the plates**:
    - The electric field due to the positive plate (left) points outward from the plate.
    - The electric field due to the negative plate (right) points inward from the plate.
    - These fields cancel each other out, resulting in \( E = 0 \).

- **(b) To the right of the plates**
Transcribed Image Text:### Electric Field between Parallel Metal Plates In the figure, two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge densities of opposite signs and magnitude \(6.51 \times 10^{-22} \, \text{C/m}^2\). The problem is to find the magnitude of the electric field at points: - (a) To the left of the plates - (b) To the right of them - (c) Between them #### Explanation of the Diagram The diagram shows two parallel plates that are thin and large relative to the distance separating them. The plates are aligned vertically, and the \( x \)-axis runs horizontally through the center of each plate. 1. **Left Plate (Positive Charge)**: - The left plate has lines indicating a positive surface charge density. 2. **Right Plate (Negative Charge)**: - The right plate has lines indicating a negative surface charge density. #### Points to Identify - **(a) To the left of the plates**: - This point is located outside and to the left of the positively charged plate. - **(b) To the right of the plates**: - This point is located outside and to the right of the negatively charged plate. - **(c) Between the plates**: - This point is located in the region directly between the positive and negative plates. ### Calculation of Electric Fields Using the principle that the electric field \( E \) due to a large, flat, uniformly charged plate with charge density \( \sigma \) is given by: \[ E = \frac{\sigma}{2\epsilon_0} \] Where: - \( \sigma \) is the surface charge density - \( \epsilon_0 \) is the permittivity of free space (\( \epsilon_0 \approx 8.854 \times 10^{-12} \, \text{Fm}^{-1} \)) Here are the specific calculations for each region: - **(a) To the left of the plates**: - The electric field due to the positive plate (left) points outward from the plate. - The electric field due to the negative plate (right) points inward from the plate. - These fields cancel each other out, resulting in \( E = 0 \). - **(b) To the right of the plates**
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