4) A small plastic ball is suspended by a 0.30 m long thread in a uniform, horizontal electric field as shown in the figure. An electric field of 1,000 V/m is directed toward the right. The ball has a mass of 2.0x10³ kg. If the ball is in equilibrium when the string makes an angle of 12° with the vertical as shown, what is the net charge on the ball? Clearly draw and label all forces and show how you arrive at your result. E=1.00 x 10' N/C L

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**Problem 4: Equilibrium of a Charged Ball in an Electric Field**

A small plastic ball is suspended by a 0.30 m long thread in a uniform, horizontal electric field as shown in the figure. An electric field of 1,000 V/m is directed toward the right. The ball has a mass of \(2.0 \times 10^{-3}\) kg. If the ball is in equilibrium when the string makes an angle of \(12^\circ\) with the vertical as shown, what is the net charge on the ball? Clearly draw and label all forces and show how you arrive at your result.

---

**Diagram Explanation:**

- The diagram illustrates a ball suspended by a thread making an angle \(\theta = 12^\circ\) with the vertical.
- The electric field \(E = 1.00 \times 10^3 \, \text{N/C}\) is directed horizontally to the right.
- Forces acting on the ball include gravitational force (\(mg\)), tension (\(T\)) in the thread, and the electric force (\(F_e\)) due to the electric field.

**Steps to Solve:**

1. **Identify Forces:**
   - Gravitational Force: \(F_g = mg\)
   - Electric Force: \(F_e = qE\)
   - Tension in the string (\(T\)) has components balancing \(F_g\) and \(F_e\):
     - Vertical component: \(T \cos \theta = mg\)
     - Horizontal component: \(T \sin \theta = qE\)

2. **Equilibrium Conditions:**
   - Vertical: \(T \cos \theta = mg\)
   - Horizontal: \(T \sin \theta = qE\)

3. **Solve for Charge (\(q\)):**
   - From \(T \cos \theta = mg\):
     \[
     T = \frac{mg}{\cos \theta}
     \]
   - Substitute into \(T \sin \theta = qE\):
     \[
     \frac{mg \sin \theta}{\cos \theta} = qE
     \]
     \[
     q = \frac{mg \tan \theta}{E}
     \]

4. **Substitute Values:**
   - \(m = 2.0 \times 10^{-3}\) kg
Transcribed Image Text:**Problem 4: Equilibrium of a Charged Ball in an Electric Field** A small plastic ball is suspended by a 0.30 m long thread in a uniform, horizontal electric field as shown in the figure. An electric field of 1,000 V/m is directed toward the right. The ball has a mass of \(2.0 \times 10^{-3}\) kg. If the ball is in equilibrium when the string makes an angle of \(12^\circ\) with the vertical as shown, what is the net charge on the ball? Clearly draw and label all forces and show how you arrive at your result. --- **Diagram Explanation:** - The diagram illustrates a ball suspended by a thread making an angle \(\theta = 12^\circ\) with the vertical. - The electric field \(E = 1.00 \times 10^3 \, \text{N/C}\) is directed horizontally to the right. - Forces acting on the ball include gravitational force (\(mg\)), tension (\(T\)) in the thread, and the electric force (\(F_e\)) due to the electric field. **Steps to Solve:** 1. **Identify Forces:** - Gravitational Force: \(F_g = mg\) - Electric Force: \(F_e = qE\) - Tension in the string (\(T\)) has components balancing \(F_g\) and \(F_e\): - Vertical component: \(T \cos \theta = mg\) - Horizontal component: \(T \sin \theta = qE\) 2. **Equilibrium Conditions:** - Vertical: \(T \cos \theta = mg\) - Horizontal: \(T \sin \theta = qE\) 3. **Solve for Charge (\(q\)):** - From \(T \cos \theta = mg\): \[ T = \frac{mg}{\cos \theta} \] - Substitute into \(T \sin \theta = qE\): \[ \frac{mg \sin \theta}{\cos \theta} = qE \] \[ q = \frac{mg \tan \theta}{E} \] 4. **Substitute Values:** - \(m = 2.0 \times 10^{-3}\) kg
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