Answer :
Let's break down the problem step-by-step:
### Given:
1. Electric field, [tex]\(\overrightarrow{E} = -6.80 \times 10^5 \, \text{N/C}\)[/tex] in the [tex]\(x\)[/tex]-direction.
2. Distance traveled by the proton before coming to rest, [tex]\( d = 6.90 \,\text{cm} = 0.069 \,\text{m} \)[/tex].
3. Proton charge, [tex]\( q = 1.60 \times 10^{-19} \, \text{C} \)[/tex].
4. Proton mass, [tex]\( m = 1.67 \times 10^{-27} \, \text{kg} \)[/tex].
### Step-by-step solution:
#### (a) Determine the acceleration of the proton.
1. Find the force on the proton due to the electric field:
[tex]\[ F = qE \][/tex]
Substituting the values:
[tex]\[ F = (1.60 \times 10^{-19} \, \text{C}) \times (-6.80 \times 10^5 \, \text{N/C}) = -1.088 \times 10^{-13} \, \text{N} \][/tex]
2. Use Newton's second law to find the acceleration ([tex]\( a \)[/tex]) of the proton:
[tex]\[ F = ma \][/tex]
[tex]\[ a = \frac{F}{m} = \frac{-1.088 \times 10^{-13} \, \text{N}}{1.67 \times 10^{-27} \, \text{kg}} \approx -6.52 \times 10^{13} \, \text{m/s}^2 \][/tex]
3. The magnitude of the acceleration is:
[tex]\[ 6.52 \times 10^{13} \, \text{m/s}^2 \][/tex]
4. The direction of the acceleration is in the negative [tex]\(x\)[/tex]-direction (opposite to the direction of the electric field).
Therefore, the acceleration of the proton is:
- Magnitude: [tex]\( 6.52 \times 10^{13} \, \text{m/s}^2 \)[/tex]
- Direction: Negative [tex]\(x\)[/tex]-direction
#### (b) Determine the initial speed of the proton.
1. Use the kinematic equation:
[tex]\[ v_f^2 = v_i^2 + 2ad \][/tex]
Since the proton comes to rest, [tex]\( v_f = 0 \)[/tex]. Therefore:
[tex]\[ 0 = v_i^2 + 2(-6.52 \times 10^{13} \, \text{m/s}^2)(0.069 \, \text{m}) \][/tex]
Solving for [tex]\( v_i \)[/tex]:
[tex]\[ v_i^2 = -2 \cdot (-6.52 \times 10^{13} \, \text{m/s}^2) \cdot 0.069 \, \text{m} \][/tex]
[tex]\[ v_i^2 = 8.99 \times 10^{12} \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ v_i = \sqrt{8.99 \times 10^{12} \, \text{m}^2/\text{s}^2} \approx 3.00 \times 10^6 \, \text{m/s} \][/tex]
2. The direction of the initial speed is in the positive [tex]\(x\)[/tex]-direction.
Therefore, the initial speed of the proton is:
- Magnitude: [tex]\( 3.00 \times 10^6 \, \text{m/s} \)[/tex]
- Direction: Positive [tex]\(x\)[/tex]-direction
#### (c) Determine the time interval over which the proton comes to rest.
1. Use the kinematic equation [tex]\( v_f = v_i + at \)[/tex]. Since [tex]\( v_f = 0 \)[/tex]:
[tex]\[ 0 = 3.00 \times 10^6 \, \text{m/s} + (-6.52 \times 10^{13} \, \text{m/s}^2)t \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{3.00 \times 10^6 \, \text{m/s}}{6.52 \times 10^{13} \, \text{m/s}^2} \approx 4.60 \times 10^{-8} \, \text{s} \][/tex]
Therefore, the time interval over which the proton comes to rest is:
- [tex]\( 4.60 \times 10^{-8} \, \text{s} \)[/tex]
To summarize, the answers are:
- (a) Acceleration: [tex]\( 6.52 \times 10^{13} \, \text{m/s}^2 \)[/tex], Direction: Negative [tex]\(x\)[/tex]-direction
- (b) Initial speed: [tex]\( 3.00 \times 10^6 \, \text{m/s} \)[/tex], Direction: Positive [tex]\(x\)[/tex]-direction
- (c) Time interval: [tex]\( 4.60 \times 10^{-8} \, \text{s} \)[/tex]
### Given:
1. Electric field, [tex]\(\overrightarrow{E} = -6.80 \times 10^5 \, \text{N/C}\)[/tex] in the [tex]\(x\)[/tex]-direction.
2. Distance traveled by the proton before coming to rest, [tex]\( d = 6.90 \,\text{cm} = 0.069 \,\text{m} \)[/tex].
3. Proton charge, [tex]\( q = 1.60 \times 10^{-19} \, \text{C} \)[/tex].
4. Proton mass, [tex]\( m = 1.67 \times 10^{-27} \, \text{kg} \)[/tex].
### Step-by-step solution:
#### (a) Determine the acceleration of the proton.
1. Find the force on the proton due to the electric field:
[tex]\[ F = qE \][/tex]
Substituting the values:
[tex]\[ F = (1.60 \times 10^{-19} \, \text{C}) \times (-6.80 \times 10^5 \, \text{N/C}) = -1.088 \times 10^{-13} \, \text{N} \][/tex]
2. Use Newton's second law to find the acceleration ([tex]\( a \)[/tex]) of the proton:
[tex]\[ F = ma \][/tex]
[tex]\[ a = \frac{F}{m} = \frac{-1.088 \times 10^{-13} \, \text{N}}{1.67 \times 10^{-27} \, \text{kg}} \approx -6.52 \times 10^{13} \, \text{m/s}^2 \][/tex]
3. The magnitude of the acceleration is:
[tex]\[ 6.52 \times 10^{13} \, \text{m/s}^2 \][/tex]
4. The direction of the acceleration is in the negative [tex]\(x\)[/tex]-direction (opposite to the direction of the electric field).
Therefore, the acceleration of the proton is:
- Magnitude: [tex]\( 6.52 \times 10^{13} \, \text{m/s}^2 \)[/tex]
- Direction: Negative [tex]\(x\)[/tex]-direction
#### (b) Determine the initial speed of the proton.
1. Use the kinematic equation:
[tex]\[ v_f^2 = v_i^2 + 2ad \][/tex]
Since the proton comes to rest, [tex]\( v_f = 0 \)[/tex]. Therefore:
[tex]\[ 0 = v_i^2 + 2(-6.52 \times 10^{13} \, \text{m/s}^2)(0.069 \, \text{m}) \][/tex]
Solving for [tex]\( v_i \)[/tex]:
[tex]\[ v_i^2 = -2 \cdot (-6.52 \times 10^{13} \, \text{m/s}^2) \cdot 0.069 \, \text{m} \][/tex]
[tex]\[ v_i^2 = 8.99 \times 10^{12} \, \text{m}^2/\text{s}^2 \][/tex]
[tex]\[ v_i = \sqrt{8.99 \times 10^{12} \, \text{m}^2/\text{s}^2} \approx 3.00 \times 10^6 \, \text{m/s} \][/tex]
2. The direction of the initial speed is in the positive [tex]\(x\)[/tex]-direction.
Therefore, the initial speed of the proton is:
- Magnitude: [tex]\( 3.00 \times 10^6 \, \text{m/s} \)[/tex]
- Direction: Positive [tex]\(x\)[/tex]-direction
#### (c) Determine the time interval over which the proton comes to rest.
1. Use the kinematic equation [tex]\( v_f = v_i + at \)[/tex]. Since [tex]\( v_f = 0 \)[/tex]:
[tex]\[ 0 = 3.00 \times 10^6 \, \text{m/s} + (-6.52 \times 10^{13} \, \text{m/s}^2)t \][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{3.00 \times 10^6 \, \text{m/s}}{6.52 \times 10^{13} \, \text{m/s}^2} \approx 4.60 \times 10^{-8} \, \text{s} \][/tex]
Therefore, the time interval over which the proton comes to rest is:
- [tex]\( 4.60 \times 10^{-8} \, \text{s} \)[/tex]
To summarize, the answers are:
- (a) Acceleration: [tex]\( 6.52 \times 10^{13} \, \text{m/s}^2 \)[/tex], Direction: Negative [tex]\(x\)[/tex]-direction
- (b) Initial speed: [tex]\( 3.00 \times 10^6 \, \text{m/s} \)[/tex], Direction: Positive [tex]\(x\)[/tex]-direction
- (c) Time interval: [tex]\( 4.60 \times 10^{-8} \, \text{s} \)[/tex]
