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Parallel Plates and Electric Field

  1. Mar 1, 2007 #1
    1. The problem statement, all variables and given/known data

    The electrostatic potential energy of a system of point charges q1 = 1 µC, q2 = 2 µC, and q3 = 3 µC at the corners of an equilateral triangle whose side s = 30 cm is


    a. 1.10 J

    b. 0.990 J

    c. 0.631 J

    d. 0.330 J

    e. 0.123 J



    2. Relevant equations

    See below.

    3. The attempt at a solution

    PE = k*q*q_0/(r)

    PE_total = [(k*(1*10^-6 C)*(3*10^-6 C)/(.30 m)] + [(k*(1*10^-6 C)*(2*10^-6 C)/(.30 m)] + [(k*(3*10^-6 C)*(2*10^-6 C)/(.30 m)]

    = k/.30 m*[(3*10^-12 C^2) + (2*10^-12) + (6*10^-12)] = 0.3296 J ??










    1. The problem statement, all variables and given/known data


    Two flat parallel plates are a distance d = 0.40 cm apart. The potential difference between the plates is 360 V. The electric field at a point midway between the plates is approximately


    a. 90 kN/C

    b. 3.6 kN/C

    c. 0.9 kN/C

    d. zero

    e. 3.6*10^5 N/C

    2. Relevant equations

    See below.

    3. The attempt at a solution

    E = delta V/d = 360 V/0.004 m = 90,000 V/m = 90,000 N/C ?

    Thanks.
     
  2. jcsd
  3. Mar 2, 2007 #2
    Your solutions appear correct to me.

    Potential energy is the sum of the energys required to add each charge in sequence (when dealing with discrete charges). You have followed this rule in your calculations adding first the 1 microC, then the 3 microC then the 2 microC (this is implicit in the way you wrote out the equations). The order is not important since E fields superpose linearly (ie they add up in the simplest way).
    The second problem also seems right to me. This is a uniform field problem and the calculated value of field strength is constant thoughout the region.
     
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