Two 1.0 (g) balls are connected by a 2.0 (cm) long insulating rod of negligible mass. One ball has a charge of +10 (nC), the other a charge of -10 (nC). The rod is held in a 1.0 × 104 (N/C) uniform electric field at an angle of 30° with respect to the field, then released. What is its initial angular acceleration? Consider I = 2.0 × 10-7 (kg m?).
Two 1.0 (g) balls are connected by a 2.0 (cm) long insulating rod of negligible mass. One ball has a charge of +10 (nC), the other a charge of -10 (nC). The rod is held in a 1.0 × 104 (N/C) uniform electric field at an angle of 30° with respect to the field, then released. What is its initial angular acceleration? Consider I = 2.0 × 10-7 (kg m?).
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
Transcribed Image Text:### Physics Problem: Angular Acceleration of Charged Balls in an Electric Field
#### Problem Statement:
Two 1.0 gram (g) balls are connected by a 2.0 centimeter (cm) long insulating rod of negligible mass. One ball has a charge of +10 nanocoulombs (nC), and the other has a charge of -10 nanocoulombs (nC). The rod is held in a uniform electric field of \(1.0 \times 10^4 \text{ N/C}\) at an angle of \(30^\circ\) with respect to the field and then released. Determine the rod's initial angular acceleration. The moment of inertia (I) of the system is given as \(2.0 \times 10^{-7} \text{ kg m}^2\).
#### Explanation:
1. **Mass and Charges**:
- **Mass of each ball**: 1.0 gram (g), which is equivalent to \(1.0 \times 10^{-3}\) kilograms (kg).
- **Charge of one ball**: +10 nanocoulombs (nC) which is \(+10 \times 10^{-9}\) coulombs (C).
- **Charge of the other ball**: -10 nanocoulombs (nC) which is \(-10 \times 10^{-9}\) coulombs (C).
2. **Rod Length**:
- **Distance between the two balls**: 2.0 centimeters (cm), which is 0.02 meters (m).
3. **Electric Field**:
- **Magnitude**: \(1.0 \times 10^4 \text{ N/C}\)
- **Angle with respect to the field**: \(30^\circ\)
4. **Moment of Inertia (I)**:
- \(2.0 \times 10^{-7} \text{ kg m}^2\)
#### Objective:
- Calculate the initial angular acceleration of the system once it is released.
#### Solution Strategy:
1. **Torque Calculation**:
- The force on each charge due to the electric field can be calculated using \( \mathbf{F} = q \mathbf{E} \).
- For the +10 nC charge: \( F_{+} = ( +10 \times 10^{-9} \text{
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