A proton p and an electron e are arranged on the x-axis as below. The direction of the electric field at points 1, 2 and 3 (respectively) are: 1 2 3 P X a) →,-, b) -,- c)-,- d) ,+ e)–,0,+

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**Electric Field Direction Problem**

A proton \( p \) and an electron \( e \) are arranged on the x-axis as shown in the diagram below. The task is to determine the direction of the electric field at points 1, 2, and 3 (respectively).

**Diagram Description:**
- The proton \( P \) is positioned at point 2 along the x-axis.
- The electron \( e \) is positioned to the right of the proton at point 3.
- Another point of interest, point 1, is located to the left of the proton along the x-axis.

**Options for Electric Field Directions:**

a) \( \rightarrow, \leftarrow, \rightarrow \)  
b) \( \leftarrow, \rightarrow, \leftarrow \)  
c) \( \leftarrow, \rightarrow, \rightarrow \)  
d) \( \leftarrow, \leftarrow, \rightarrow \)  
e) \( \leftarrow, 0, \leftarrow \)

**Answer Choices:**

- \( \bigcirc \) b  
- \( \bigcirc \) a  
- \( \bigcirc \) c  
- \( \bigcirc \) d  
- \( \bigcirc \) e  

**Explanation:**

In this setup:

- At point 1, which is to the left of the proton \( P \), the electric field created by the proton points away from it (\( \rightarrow \)), and the electric field created by the electron at point 3 (farther to the right) will point towards it (\( \leftarrow \)). The proton's influence is more significant at this closer point, resulting in a net electric field direction towards the proton, hence \( \leftarrow \).
  
- At point 2, directly on the proton, the proton's electric field would be zero because it is at the location of the source charge. However, due to the presence of the electron \( e \) to the right, the electric field at this point would point towards the negatively charged electron, hence \( \rightarrow \).

- At point 3, which is to the right of the electron \( e \), the electric field created by the proton (pointing away from the proton \( \rightarrow \)) and the electric field by the electron (pointing towards the electron \( \leftarrow \)). The electron's influence at this closer point results in a net electric field
Transcribed Image Text:**Electric Field Direction Problem** A proton \( p \) and an electron \( e \) are arranged on the x-axis as shown in the diagram below. The task is to determine the direction of the electric field at points 1, 2, and 3 (respectively). **Diagram Description:** - The proton \( P \) is positioned at point 2 along the x-axis. - The electron \( e \) is positioned to the right of the proton at point 3. - Another point of interest, point 1, is located to the left of the proton along the x-axis. **Options for Electric Field Directions:** a) \( \rightarrow, \leftarrow, \rightarrow \) b) \( \leftarrow, \rightarrow, \leftarrow \) c) \( \leftarrow, \rightarrow, \rightarrow \) d) \( \leftarrow, \leftarrow, \rightarrow \) e) \( \leftarrow, 0, \leftarrow \) **Answer Choices:** - \( \bigcirc \) b - \( \bigcirc \) a - \( \bigcirc \) c - \( \bigcirc \) d - \( \bigcirc \) e **Explanation:** In this setup: - At point 1, which is to the left of the proton \( P \), the electric field created by the proton points away from it (\( \rightarrow \)), and the electric field created by the electron at point 3 (farther to the right) will point towards it (\( \leftarrow \)). The proton's influence is more significant at this closer point, resulting in a net electric field direction towards the proton, hence \( \leftarrow \). - At point 2, directly on the proton, the proton's electric field would be zero because it is at the location of the source charge. However, due to the presence of the electron \( e \) to the right, the electric field at this point would point towards the negatively charged electron, hence \( \rightarrow \). - At point 3, which is to the right of the electron \( e \), the electric field created by the proton (pointing away from the proton \( \rightarrow \)) and the electric field by the electron (pointing towards the electron \( \leftarrow \)). The electron's influence at this closer point results in a net electric field
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