A proton is traveling horizontally to the right at 4.70 x 10^6 m/s. Find the direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.50 cm. 

To find the direction of the electric field that can bring the proton uniformly to rest, we need to consider how the electric field interacts with the proton. Since the proton is positively charged, it will experience a force in the direction of the electric field. The electric field must decelerate the proton, so it must exert a force opposite to the proton's direction of motion.

### Steps to Solve:

1. **Determine the deceleration required**:
   Using the kinematic equation for uniform acceleration:
   \[
   v_f^2 = v_i^2 + 2a d
   \]
   where:
   - \( v_f = 0 \, \text{m/s} \) (final velocity, because the proton comes to rest),
   - \( v_i = 4.70 \times 10^6 \, \text{m/s} \) (initial velocity),
   - \( d = 3.50 \, \text{cm} = 0.035 \, \text{m} \) (distance over which the proton comes to rest),
   - \( a \) is the acceleration.

   Solving for \( a \):
   \[
   0 = (4.70 \times 10^6)^2 + 2a(0.035)
   \]
   \[
   a = - \frac{(4.70 \times 10^6)^2}{2 \times 0.035}
   \]
   \[
   a \approx -3.16 \times 10^{14} \, \text{m/s}^2
   \]
   (The negative sign indicates that the acceleration is in the opposite direction of the proton's motion.)

2. **Relate the acceleration to the electric field**:
   The force on the proton due to the electric field is given by \( F = qE \), where \( q \) is the charge of the proton and \( E \) is the electric field strength. Using Newton's second law \( F = ma \), we get:
   \[
   qE = ma
   \]
   Solving for \( E \):
   \[
   E = \frac{ma}{q}
   \]
   where:
   - \( m = 1.67 \times 10^{-27} \, \text{kg} \) (mass of the proton),
   - \( a = -3.16 \times 10^{14} \, \text{m/s}^2 \) (deceleration),
   - \( q = 1.60 \times 10^{-19} \, \text{C} \) (charge of the proton).

   Substituting the values:
   \[
   E = \frac{(1.67 \times 10^{-27})(-3.16 \times 10^{14})}{1.60 \times 10^{-19}}
   \]
   \[
   E \approx -3.30 \times 10^3 \, \text{N/C}
   \]
   The negative sign indicates that the electric field is in the opposite direction of the proton's initial velocity.

### Conclusion:
- The **weakest electric field** that can bring the proton to rest over the given distance is approximately **\( 3.30 \times 10^3 \, \text{N/C} \)**.
- The **direction of the electric field** is **to the left**, opposite to the direction of the proton's motion.

Would you like further clarification or additional details?

Here are 5 related questions:
1. How would the problem change if the proton had a different initial velocity?
2. What would happen if the distance to stop the proton was doubled?
3. How does the charge of the proton affect the required electric field?
4. What would the required electric field be for a particle with a negative charge, like an electron?
5. Can you describe how an electric field can change the motion of a charged particle?

**Tip**: Always pay attention to units when solving problems with different quantities like distance, charge, and mass. Converting units to standard SI units early on can help avoid errors later in the calculation.

A proton is traveling horizontally to the right at 4.70 x 10^6 m/s. Find the direction of the weakest electric field that can bring the proton uniformly to rest over a distance of 3.50 cm.

Learn how to calculate the weakest electric field that can uniformly bring a proton traveling at 4.70 x 10^6 m/s to rest over a distance of 3.50 cm. This problem involves kinematic equations and electrostatics, suitable for undergraduate physics students.

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