4.A plastic rod with a uniform charge -Q is bent in a 120° circular arc of radius r and symmetrically paced across an x axis with the origin at the center of curvature P of the rod. In terms of Q and r, what is the electric field E due to the rod at point P? ( E as a function of K, Q and r)
4.A plastic rod with a uniform charge -Q is bent in a 120° circular arc of radius r and symmetrically paced across an x axis with the origin at the center of curvature P of the rod. In terms of Q and r, what is the electric field E due to the rod at point P? ( E as a function of K, Q and r)
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Chapter1: Units, Trigonometry. And Vectors
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![### Question 4: Electric Field Due to a Charged Rod
**Problem Statement:**
A plastic rod with a uniform charge -Q is bent in a 120° circular arc of radius \( r \) and symmetrically placed across an x-axis with the origin at the center of curvature \( P \) of the rod. In terms of \( Q \) and \( r \), what is the electric field \( \vec{E} \) due to the rod at point \( P \)? (Express \( \vec{E} \) as a function of \( K \), \( Q \), and \( r \)).
**Explanation:**
- A plastic rod is uniformly charged with a negative charge \(-Q\).
- The rod is shaped into a circular arc covering 120 degrees.
- The radius of this arc is \( r \).
- The arc is positioned such that it is symmetric around the x-axis.
- The origin (\( O \)) is at the center of curvature of the arc.
- We aim to find the electric field \( \vec{E} \) at point \( P \), which is at the center of curvature, in terms of the constants \( K \), \( Q \), and \( r \).
#### Notable Variables:
- \( Q \): Total charge on the rod.
- \( r \): Radius of the circular arc.
- \( K \): Coulomb's constant.
**Detailed Solution**
Given that the charge is uniformly distributed along the arc, we can calculate the linear charge density \( \lambda \) (charge per unit length):
\[ \lambda = \frac{Q}{L} \]
where \( L \) is the length of the arc. The length of the arc can be calculated as:
\[ L = r \cdot \theta \]
where \( \theta \) is the angle in radians. For a 120° arc, we convert degrees to radians:
\[ \theta = \frac{120^\circ}{180^\circ} \pi = \frac{2\pi}{3} \]
Thus:
\[ L = r \cdot \frac{2\pi}{3} = \frac{2\pi r}{3} \]
Therefore, the linear charge density \( \lambda \) is:
\[ \lambda = \frac{-Q}{\frac{2\pi r}{3}} = \frac{-3Q}{2\pi](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2d1ccfb0-37c4-4041-bafd-68224be01209%2F191889ef-ecf2-460c-aa6d-b2c09d79c4db%2Fsz32vfy_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 4: Electric Field Due to a Charged Rod
**Problem Statement:**
A plastic rod with a uniform charge -Q is bent in a 120° circular arc of radius \( r \) and symmetrically placed across an x-axis with the origin at the center of curvature \( P \) of the rod. In terms of \( Q \) and \( r \), what is the electric field \( \vec{E} \) due to the rod at point \( P \)? (Express \( \vec{E} \) as a function of \( K \), \( Q \), and \( r \)).
**Explanation:**
- A plastic rod is uniformly charged with a negative charge \(-Q\).
- The rod is shaped into a circular arc covering 120 degrees.
- The radius of this arc is \( r \).
- The arc is positioned such that it is symmetric around the x-axis.
- The origin (\( O \)) is at the center of curvature of the arc.
- We aim to find the electric field \( \vec{E} \) at point \( P \), which is at the center of curvature, in terms of the constants \( K \), \( Q \), and \( r \).
#### Notable Variables:
- \( Q \): Total charge on the rod.
- \( r \): Radius of the circular arc.
- \( K \): Coulomb's constant.
**Detailed Solution**
Given that the charge is uniformly distributed along the arc, we can calculate the linear charge density \( \lambda \) (charge per unit length):
\[ \lambda = \frac{Q}{L} \]
where \( L \) is the length of the arc. The length of the arc can be calculated as:
\[ L = r \cdot \theta \]
where \( \theta \) is the angle in radians. For a 120° arc, we convert degrees to radians:
\[ \theta = \frac{120^\circ}{180^\circ} \pi = \frac{2\pi}{3} \]
Thus:
\[ L = r \cdot \frac{2\pi}{3} = \frac{2\pi r}{3} \]
Therefore, the linear charge density \( \lambda \) is:
\[ \lambda = \frac{-Q}{\frac{2\pi r}{3}} = \frac{-3Q}{2\pi
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