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Electric field strength and electrostatic potential

  1. Aug 20, 2013 #1
    1. The problem is in the attachments.



    2. Field strength is equal to voltage divided by distance



    3. I want to make sure that i understand a few things before i solve this problem! i would like to know your opinion!

    a) first, field strength is inversely proportional to the the distance squared that means filed strength decreases with distance, and as the probe approaches the planet that must mean that the charge is not coming from the planet, it's coming from behind the probe!

    b) the negative sign here means that the charge is negative

    c) i'm not really sure about this so plz could u shed some light on it, i want to know what happens to the voltage as the distance increases from the charge, ite decreases correct? and what would the negative charge of the voltage difference mean?! excess in negative charge of the point closer to the charge ?!

    d) my attempt at the solution, the field strength has to be uniform so i take the average which is 500+400/2= -450, then v = field strength times distance which is 20,000,000 meters = -9000MV
     

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    Last edited: Aug 20, 2013
  2. jcsd
  3. Aug 20, 2013 #2
    looks right to me...
     
  4. Aug 20, 2013 #3

    rude man

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    Is the sign right? What's the expression for ΔV given E?
     
  5. Aug 20, 2013 #4
    i don't know an expression for delta v, plz tell me!
     
  6. Aug 21, 2013 #5

    rude man

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    Think of a parallel plate capacitor. The top plate is + and the bottom is -. A positive test charge is moved from the bottom to the top plate. Call s = 0 at the bottom plate and s = d at the top plate where d is the distance between the plates.

    What is the direction of the E field? What is its sign? What is the difference in potential as we move the test charge from the bottom to the top plate?

    The formal expression is ΔV = Vtop - Vbottom = -∫E*ds. So you wind up with what sign for the potential difference going from the bottom to the top plate?

    (Vectors in bold. * denotes dot-product).
     
  7. Aug 21, 2013 #6
    I never studied capacitors before, and i never took calculus :(
     
  8. Aug 21, 2013 #7
    plz help!
     
  9. Aug 21, 2013 #8
    guys my exam is on the 25th i need an answer!
     
  10. Aug 22, 2013 #9
    omg no one? :(
     
  11. Aug 22, 2013 #10

    rude man

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    I can't speak for everyone but I feel helpless to give you any more hints in view of you limited background.

    Do you know what potential energy is? Do you know what happens to potential energy of a positive charge if you move that charge along an electric field that opposes the motion?
     
  12. Aug 22, 2013 #11

    yes, it has PE equal to kqq/d, plz help me understand i have a very important exam :(
     
  13. Aug 23, 2013 #12

    rude man

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    Well, the change in potential equals the change in potential energy of a unit positive charge. And the change in potential energy of a unit positive charge is the negative of the product of the E field times the distance covered.

    So what does that make the change in potential if E is negative? Plus or minus?
     
  14. Aug 23, 2013 #13
    well that makes it a plus but i still don't understand why! can you plz elaborate in some more detail, perhaps take some time to answer some of my questions above? would u like to discuss this on skype? my exam is on 25th, this is so urgent :(
     
  15. Aug 23, 2013 #14

    rude man

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    That makes it a plus.
     
  16. Aug 23, 2013 #15
    thx for your thorough explanation rude man, if this problem came in my exam i will choose minus, waiting for a more clear explanation from someone who is more generous
     
  17. Aug 23, 2013 #16

    rude man

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    Ok, sorry to disappoint you. Let's try this:

    Since the magnitude of the E field is diminishing as the spaceship approaches Earth, the source of the E field must be behind it. That's because |E| = k|Q|/r^2 where Q is the source of the E field and r is the distance between Q and the spaceship.

    And since the E field is negative, this means Q, the source of the E field, must be a negative charge: E = -kQ/r^2. r is measured from the position of the source of charge to the spaceship and is always positive, directed away from the charge.

    So, as the spaceship travels, r increases and the potential, which is kQ/r, increases, i.e. goes more positive. Remember, Q is negative so as r gets bigger, the magnitude of the potential gets smaller which means the potential gets more positive.

    So the change in potential is positive.
     
  18. Aug 23, 2013 #17
    first thank you very much for taking the time to explain, i'm new to this forum and i noticed ppl here (helpers) like to keep the answers as short as possible! secondly, i fully understand what u said...however, my text book says if the field and two points are like this: E------point 1------point 2 then the potential difference is v at point 1 - v at point 2, so negative - positive will result in negative! your opinion?
     
  19. Aug 23, 2013 #18

    rude man

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    V at point 1 is kQ/r1. V at point 2 is kQ/r2. So if r2 > r1 and Q is negative, V2 - V1 is positive.
     
  20. Aug 23, 2013 #19
    i don't think u understand negative here, well it says that field was -500 then decreased to - 450, the negative as u know means due to negative charge, negative voltage means excess in negative charge, v at 1 is bigger than v at 2, the negative is not involved in the math!
     
  21. Aug 24, 2013 #20
    time's up, exam tomorrow :(
     
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