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IIT JEE problem (capacitors)

  1. Jun 5, 2016 #1
    1. The problem statement, all variables and given/known data
    upload_2016-6-6_9-27-53.png

    2. Relevant equations


    3. The attempt at a solution
    1)
    the first question is easy,it is (C) because these charged balls are repelled from the similarly charged plates and accelerate towards the negatively charged plate,and once thy stick to the second plate they transfer double the charge they initially had (to the plate) and become oppositely charged.
    the same thing happens again and again in a periodic way(which is in no way simple harmonic).
    this discrete charge transfer at particular instants of time causes a current to flow(discrete current).

    2)
    the second one i got wrong,i do not know why.
    i assumed that the average current is proportional to the velocity of the balls,i can relate the velocities to the potential applied as follows
    (velocity)^2 proportional to (potential diff)
    this is directly from the work energy theorem
    now (velocity) proportional to (potential diff)^1/2
    from this and the previous assumption we see that (avg current) proportional to (potential diff)^1/2
    and so the correct answer is (C) but in their answer keys it is given the answer is (D) (ie proportional to V^2)
    i don't see how this can be.
     
  2. jcsd
  3. Jun 6, 2016 #2

    NascentOxygen

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    Staff: Mentor

    Can you explain in detail this part.
     
  4. Jun 6, 2016 #3
    hello nascent oxygen
    it follows from the work energy theorem
    KE=1/2mv^2=work done on the ball
    but the work done on the ball=potential through which it is accelerated*charge on ball
    so 1/2mv^2=qV
    here V= potential
    v= velocity
    we see v^2 proportional to V
     
  5. Jun 6, 2016 #4
    hello nascent oxygen
    it follows from the work energy theorem
    KE=1/2mv^2=work done on the ball
    but the work done on the ball=potential through which it is accelerated*charge on ball
    so 1/2mv^2=qV
    here V= potential
    v= velocity
    we see v^2 proportional to V
     
  6. Jun 6, 2016 #5

    NascentOxygen

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    Staff: Mentor

    That's the final velocity.

    Then there's q. The final velocity also involves q.
     
  7. Jun 6, 2016 #6
    yes but q is a constant.
    or does it also depend on V
    oh wait it does, as you increase V charge on the balls also increases.
    but how do i know the relation between them?
     
  8. Jun 6, 2016 #7

    NascentOxygen

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    Staff: Mentor

    You're looking for the relation between a sphere's potential and the charge it carries. Try google

    As each conductive sphere contacts a plate it will take on the potential of that plate.
     
  9. Jun 6, 2016 #8
    how do you know that the sphere attains the same potential as the plate, it might fly off to the other plate well before it gets charged to the same potential.

    ok assuming that it does get charged to same potential then V(potential of sphere) = k*(charge on sphere)/(radius of sphere) .
    because all charge resides on the surface(metal sphere).
    now (charge on sphere) proportional to (potential diff between plates)
    so (v^2) proportional to (V^2) . (because of the product qV,and q being proportional to V)
    this means v proportional to V.
    but the given answer is according to the key is (D).
     
  10. Jun 6, 2016 #9
    just so you know the answer key given to us isn't final and is subject to change.
    but im pretty sure the answer keys are right
     
  11. Jun 6, 2016 #10

    NascentOxygen

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    Staff: Mentor

    Okay so far, but the question doesn't ask how speed of the balls is related to voltage.
     
    Last edited: Jun 6, 2016
  12. Jun 6, 2016 #11
    but i have assumed that the current is proportional to the average velocity of the spheres.
     
  13. Jun 6, 2016 #12
    i think i made a mistake ,the velocities i considered weren't the average they were the final velocities.
    the average velocity will be the (distance bw plates)/(time taken to go that distance)
    since the field is constant in parallel plate capacitor, the acceleration is also constant.
    so v avg =1/2*(Eq/m)t (using the constant accleleration kinematics equation s= ut+1/2*at^2 and dividing with t to get s/t)

    here t is time taken to go from one plate to the other.

    now E is proportional to V,and q is proportional to V(assuming that the spheres are charged completely to the plates potential)
    so v avg is proportional to V^2.
    i think this is right
     
  14. Jun 6, 2016 #13

    TSny

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    Homework Helper
    Gold Member

    What about the factor of t in vavg =1/2*(Eq/m)t? Does t depend on V?

    OK.
    Yes, but what else is the current proportional to?
     
  15. Jun 6, 2016 #14

    NascentOxygen

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    Staff: Mentor

    In motion with constant acceleration, how is vav related to vf?
     
  16. Jun 7, 2016 #15
    Hey everyone, I am thinking something completely different and it is quite possible that I am wrong, so forgive me if I turn out to be wrong. The way I am thinking this problem to be is that, since it is given that balls attain same potential as the plate (as they had a metal film), so from potential I got charge on each spherical shell (note that they have metal film for a reason, so we can't say that they are spheres whenever we use potential or anything like that). Exact equation does not matter as what is asked is proportionality, so charge is directly proportional to V. Now instead of thinking how to calculate proportionality between velocity and potential, I took a different route. I used analogy between (more like my brain did it on its own and I realised what I did later) current and velocity and charge and displacement/distance (to be used as situation presents). Average speed = Total distance travelled/ total time taken, similarly I used average current = total charge passed/ total time taken. We got numerator already (no we don't need no. of balls, proportionality remember!), denominator can be calculated by using F= ma = Eq, where E is electric field inside the capacitor and m is mass of ball (don't worry about it....... proportionality!). Use this to find acceleration to find time taken by using h=(1/2)*at^2.

    Again don't bother with multiple times the ball passes up and down, you will just have to multiply numerator and denominator with same variable to take it into account. With this process I got avg current proportional to V^2
     
  17. Jun 8, 2016 #16
    to clarify q proportional to V
    and is it that t proportionat to 1/V?
    from h = 1/2*(Eq/m)t^2
     
  18. Jun 8, 2016 #17
    i used integration(to find averages)
    and found that
    v avg =v initial +1/2*a*(t initial +t final)
    or else a = 2*(v avg - v inital)/(t initial +t final)
     
  19. Jun 8, 2016 #18
    since the initial velocity is zero and acceleration uniform v avg =v final/2
    they are directly proportional!!
     
  20. Jun 8, 2016 #19
    intuition says t depends on V.
    and the current is also proportional to the charge on the spheres
     
  21. Jun 8, 2016 #20
    let me try answering this question again
    the (average current) proportional to (average velocity)(charge that each carry) {the average velocity has a linear relation to the final velocity from post #18}
    we knoe that the final kinetic energy is equal to the work done on the spheres by the field.
    so 1/2mv^2 = Ed = qV
    now we also know that q proportional to V ------------------(2)
    so v proportional to V ----------------(1)
    therefore average current proportional to V^2
    im pretty sure this method is right.
     
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