2.2 Consider a system where a point charge Q with mass m hangs from a massless string (of length L), as shown in Figure 2. A uniform line of charge (with linear charge density X), attached firmly at its end points, is placed vertically, a distance D from the point where the string is attached. If Q and A are like charges, the point charge is repelled, with the string making an angle of 0 with respect to vertical. We approximate the electric field produced by the uniformly charged line to that due to an infinite line (Eq. 3). (m← L r D Figure 2: The experimental set-up to study the interaction between a line charge and a point charge. Sketch a free-body diagram for the set-up shown in Figure 2, labeling all appropriate forces. By considering the horizontal and vertical components of the forces on the point charge, prove the following expression: (4) QX 2π€0(D+ Lsine) = = (mg)tano

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It is not graded it is just the equation number.

## 2.2 Experimental Set-Up: Interaction between a Line Charge and a Point Charge

Consider a system where a point charge \( Q \) with mass \( m \) hangs from a massless string (of length \( L \)), as shown in Figure 2. A uniform line of charge (with linear charge density \( \lambda \)), attached firmly at its end points, is placed vertically, a distance \( D \) from the point where the string is attached. If \( Q \) and \( \lambda \) are like charges, the point charge is repelled, with the string making an angle of \( \theta \) with respect to vertical. We approximate the electric field produced by the uniformly charged line to that due to an infinite line (Eq. 3).

### Figure 2: Description
The diagram illustrates the experimental set-up used to study the interaction between a line charge and a point charge. It consists of:

- A vertical line charge at a distance \( D \) from the attachment point of the string.
- A point charge \( Q \) suspended by a string of length \( L \) that forms an angle \( \theta \) with the vertical.
- The repulsion between like charges causes the string to deviate from the vertical by \( \theta \). 
- Horizontal and vertical distances marked as \( r \) and \( L \sin \theta \) respectively.

### Task: Sketch and Analysis

**Sketch** a free-body diagram for the set-up shown in Figure 2, labeling all appropriate forces. By considering the horizontal and vertical components of the forces on the point charge, **prove** the following expression:

\[
\frac{Q\lambda}{2\pi \epsilon_0 (D + L \sin \theta)} = (mg) \tan \theta \quad (4)
\]

This involves balancing the gravitational force and the electric force to derive the given expression, considering the orientation and effect of the forces on the point charge.
Transcribed Image Text:## 2.2 Experimental Set-Up: Interaction between a Line Charge and a Point Charge Consider a system where a point charge \( Q \) with mass \( m \) hangs from a massless string (of length \( L \)), as shown in Figure 2. A uniform line of charge (with linear charge density \( \lambda \)), attached firmly at its end points, is placed vertically, a distance \( D \) from the point where the string is attached. If \( Q \) and \( \lambda \) are like charges, the point charge is repelled, with the string making an angle of \( \theta \) with respect to vertical. We approximate the electric field produced by the uniformly charged line to that due to an infinite line (Eq. 3). ### Figure 2: Description The diagram illustrates the experimental set-up used to study the interaction between a line charge and a point charge. It consists of: - A vertical line charge at a distance \( D \) from the attachment point of the string. - A point charge \( Q \) suspended by a string of length \( L \) that forms an angle \( \theta \) with the vertical. - The repulsion between like charges causes the string to deviate from the vertical by \( \theta \). - Horizontal and vertical distances marked as \( r \) and \( L \sin \theta \) respectively. ### Task: Sketch and Analysis **Sketch** a free-body diagram for the set-up shown in Figure 2, labeling all appropriate forces. By considering the horizontal and vertical components of the forces on the point charge, **prove** the following expression: \[ \frac{Q\lambda}{2\pi \epsilon_0 (D + L \sin \theta)} = (mg) \tan \theta \quad (4) \] This involves balancing the gravitational force and the electric force to derive the given expression, considering the orientation and effect of the forces on the point charge.
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