Results 1 to 8 of 8

Thread: Physics: Electric Fields and Projectile Motion

  1. #1

    Physics: Electric Fields and Projectile Motion

    Hi,
    I required some help with the question below. I was quite stumped when i reached this part of the question. I'm not sure sure where to start.

    QUESTION:
    Two horizontal parallel metal plates are separated by 5cm have a voltage of 2.56V applied between them so that the upper plate is positively charged. A proton is fired horizontally into the space between the plates, just below the upper plate. Assume the plates are large enough that the proton will remain in . the space and ignore any gravitational effects.

    Data: mass of proton = 1.67 x 10^-27 kg
    Charge of proton = 1.60 x 10^-19 C

    e) Identify the horizontal velocity of the proton at prior to hitting the lower plate.

    Data I have found from the other sections of the question:
    Electric Field Strength: 51.2 NC^-1
    acceleration: 4.91 x 10^9 m/s^2 down
    time to reach lower plate: 4.52 x 10^-6 sec or 4.52µs
    Vertical velocity: 2.21 x 10^4 m/s

    All I know is a=0 in the horizontal.
    Last edited by Subhotosh Khan; 01-11-2019 at 11:12 AM. Reason: typo

  2. #2
    Elite Member
    Join Date
    Jun 2007
    Posts
    18,005
    Quote Originally Posted by miaellenaylee View Post
    Hi,
    I required some help with the question below. I was quite stumped when i reached this part of the question. I'm not sure sure where to start.

    QUESTION:
    Two horizontal parallel metal plates are separated by 5cm have a voltage of 2.56V applied between them so that the upper plate is positively charged. A proton is fired horizontally into the space between the plates, just below the upper plate. Assume the plates are large enough that the proton will remain in . the space and ignore any gravitational effects.

    Data: mass of proton = 1.67 x 10^-27 kg
    Charge of proton = 1.60 x 10^-19 C

    e) Identify the horizontal velocity of the proton at prior to hitting the lower plate.

    Data i have found from the other sections of the question:
    Electric Field Strength: 51.2 NC^-1
    acceleration: 4.91 x 10^9 m/s^2 down
    time to reach lower plate: 4.52 x 10^-6 sec or 4.52µs
    Vertical velocity: 2.21 x 10^4 m/s

    All I know is a=0 in the horizontal.
    This one I am going leave for Dan (topsquark).......
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
    Elite Member
    Join Date
    Jun 2007
    Posts
    18,005
    Quote Originally Posted by Subhotosh Khan View Post
    This one I am going leave for Dan (topsquark).......
    Bump!!
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  4. #4
    Junior Member
    Join Date
    Aug 2012
    Location
    Florian
    Posts
    157
    Quote Originally Posted by miaellenaylee View Post
    Hi,
    I required some help with the question below. I was quite stumped when i reached this part of the question. I'm not sure sure where to start.

    QUESTION:
    Two horizontal parallel metal plates are separated by 5cm have a voltage of 2.56V applied between them so that the upper plate is positively charged. A proton is fired horizontally into the space between the plates, just below the upper plate. Assume the plates are large enough that the proton will remain in . the space and ignore any gravitational effects.

    Data: mass of proton = 1.67 x 10^-27 kg
    Charge of proton = 1.60 x 10^-19 C

    e) Identify the horizontal velocity of the proton at prior to hitting the lower plate.

    Data I have found from the other sections of the question:
    Electric Field Strength: 51.2 NC^-1
    acceleration: 4.91 x 10^9 m/s^2 down
    time to reach lower plate: 4.52 x 10^-6 sec or 4.52µs
    Vertical velocity: 2.21 x 10^4 m/s

    All I know is a=0 in the horizontal.
    The electric field in between the capacitor plates only has a vertical component. Thus the proton is not accelerated in the horizontal direction thus the horizontal component of the velocity of the proton is constant.

    So to find the horizontal component of the velocity we need only know what it was when the proton was first projected between the plates. We don't have the information listed to figure that out. Is this the whole problem statement?

    Also, what do you mean by "vertical velocity?" The proton accelerates toward the negative plate and thus the velocity changes. Is this, perhaps, the speed when the proton reaches the negative plate?

    Please give us the whole question.

    -Dan
    It should be possible to explain the laws of Physics to a barmaid. - Albert Einstein

  5. #5
    Junior Member
    Join Date
    Aug 2012
    Location
    Florian
    Posts
    157
    Quote Originally Posted by Subhotosh Khan View Post
    Bump!!
    Bumpity bumplty bumpity! Wheeeee!

    -Dan
    It should be possible to explain the laws of Physics to a barmaid. - Albert Einstein

  6. #6
    Quote Originally Posted by topsquark View Post
    The electric field in between the capacitor plates only has a vertical component. Thus the proton is not accelerated in the horizontal direction thus the horizontal component of the velocity of the proton is constant.

    So to find the horizontal component of the velocity we need only know what it was when the proton was first projected between the plates. We don't have the information listed to figure that out. Is this the whole problem statement?

    Also, what do you mean by "vertical velocity?" The proton accelerates toward the negative plate and thus the velocity changes. Is this, perhaps, the speed when the proton reaches the negative plate?

    Please give us the whole question.

    -Dan
    Screen Shot 2019-01-12 at 3.27.36 pm.jpg

    For this particular question, I think the vertical velocity is the at speed the proton is travelling at prior to hitting or reaching the plate below so yes your interpretation is correct. I have attached the entire question as an image, I hope you can viewed it. Part E is the section I required help with.

  7. #7
    Elite Member
    Join Date
    Jun 2007
    Posts
    18,005
    Quote Originally Posted by miaellenaylee View Post
    Screen Shot 2019-01-12 at 3.27.36 pm.jpg

    For this particular question, I think the vertical velocity is the at speed the proton is travelling at prior to hitting or reaching the plate below so yes your interpretation is correct. I have attached the entire question as an image, I hope you can viewed it. Part E is the section I required help with.


    Have you done any work with the parts (a), (b), etc.? If you did - please share your work - so that we know where to begin to help you!!
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  8. #8
    Quote Originally Posted by Subhotosh Khan View Post


    Have you done any work with the parts (a), (b), etc.? If you did - please share your work - so that we know where to begin to help you!!
    IMG_1431.jpg
    Here's all the working out I have done for the question but I'm not too sure if they are correct...

Tags for this Thread

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •