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
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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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24227ad7-0cee-4cb3-bf8f-36b2c099999b%2Fa9a38a14-a7b2-461e-a33e-05279b44cae7%2Fpkoqvd_processed.jpeg&w=3840&q=75)
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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