How Do You Calculate Proton Acceleration in an Electric Field?

[bsharif]

College home work ..please help ASAP! :)

1. A proton accelerates from rest in a uniform electric field of 639 n/c. At some time later , its speed is 1.78 * 10^6 m/s.

What is the magnitude of the acceleration of the proton ? Answer in units of m/s^2.

2. How long does it take the proton to reach this speed ? answer in units of us.

3. How far has it moved in this time interval ?
Answer in units of m

4. What is the kinetic energy at the later time ?
Answer in units of J.



Please Help ..

Thank you

 

[Gokul43201]

https://www.physicsforums.com/showthread.php?t=4825

 

[quicknote]

for reaching out for assistance with your college homework. I am happy to help you with these questions.

1. To calculate the magnitude of acceleration, we can use the equation a = qE/m, where q is the charge of the proton (1.6 * 10^-19 C) and m is its mass (1.67 * 10^-27 kg). Plugging in the given values, we get a = (1.6 * 10^-19 C)(639 N/C)/(1.67 * 10^-27 kg) = 9.6 * 10^7 m/s^2. Therefore, the magnitude of acceleration is 9.6 * 10^7 m/s^2.

2. To find the time it takes for the proton to reach this speed, we can use the equation v = at, where v is the final velocity (1.78 * 10^6 m/s) and a is the acceleration we calculated in the previous question. So t = v/a = (1.78 * 10^6 m/s)/(9.6 * 10^7 m/s^2) = 0.0186 seconds. Therefore, it takes the proton 0.0186 seconds to reach this speed.

3. To find the distance the proton has moved in this time interval, we can use the equation d = 1/2at^2, where a is the acceleration and t is the time. Plugging in the values, we get d = 1/2(9.6 * 10^7 m/s^2)(0.0186 s)^2 = 16.6 meters. Therefore, the proton has moved 16.6 meters in this time interval.

4. The kinetic energy of the proton can be calculated using the equation KE = 1/2mv^2, where m is the mass of the proton and v is its velocity. Plugging in the values, we get KE = 1/2(1.67 * 10^-27 kg)(1.78 * 10^6 m/s)^2 = 2.5 * 10^-11 J. Therefore, the kinetic energy of the proton at the later time is 2.5 * 10^-11 J.

I hope this helps you to better understand and solve these questions. If you have any further questions or need clarification, please let me know.