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Confused about derivation of E=(V-IR)/L, multiple choice problem

  1. Jan 7, 2017 #1
    1. The problem statement, all variables and given/known data
    A vacuum photodiode detector utilizes the photoelectric effect to detect light. The photoelectric effect causes electrons to be ejected from a metal plate when photons of light are absorbed by the metal. The energy of a photon is given by the equation E = hf, where h = 6.6 x 10-34J·s (Planck’s constant), and f is the frequency of the photon. To free an electron, the energy of a photon must be greater than a quantity called the work function of the metal. The ejected electron will have a kinetic energy equal to the photon’s energy minus the work function.

    A vacuum photodiode is constructed by sealing two electrodes, a cathode and an anode, in a vacuum tube. The electrodes are separated by a distance, L = 0.01 m, and connected to a battery and a resistor, R = 100 Ω, as shown in Figure 1. The cathode is made of a photoelectric metal and is connected to the negative terminal of the battery. The potential difference between the cathode and anode is approximately equal to the battery voltage, V = 50 V. The electric field at all points between the electrodes is equal to the electrode voltage difference divided by L. The potential energy of an electron immediately after it is released from the cathode is equal to qV, where q = -1.6 x 10-19C is the charge of an electron. The work function for the vacuum photodiode is 2 x 10-19J.

    Which of the following changes to the circuit will decrease the electric field between the electrodes by the greatest amount?
    a) Increasing L by a factor of 2
    b) Decreasing L by a factor of 2
    c) Increasing R by a factor of 2
    d) Decreasing R by a factor of 2

    2. Relevant equations
    E = V/L

    3. The attempt at a solution
    FYI, there are 5 questions attached to this question stem, I'm just stuck on this first part.
    So if L = 01m and V = 50V then Einitial = V/L = 50V/.01m = 5000N/C
    also, I = 50V/100Ω = 0.5A

    a) E when increasing L by factor of 2 = 50V/(.01m*2) = 2500 N/C (decreases by 2500 N/C)
    b) E when decreasing L by factor of 2 = 50V/(.01m/2) = 10000 N/C (increases by 5000 N/C)
    c) E when increasing R by factor of 2 = (0.5A*200Ω) / .01m = 10000 N/C (increases by 5000 N/C)
    d) E when decreasing R by factor of 2 = (0.5A*50Ω) / .01m = 2500 N/C (decreases by 2500 N/C)

    The answer is (a) , but I cant figure out why answer (d) also gives a 2500 decrease. I am obviously doing something wrong.
    The answer hint states that E= (V-IR)/L but I'm confused how they get that. Thanks for any help
  2. jcsd
  3. Jan 7, 2017 #2


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    How can 0.5A flow in the circuit? How does the current get across the gap between the cathode and the anode?
  4. Jan 7, 2017 #3
    Thanks for the reply phyzguy. When the electron gets ejected from the cathode, that = the current getting across the gap. So (if that's true) wouldn't that mean that there couldn't be a 0.5A current flowing?
  5. Jan 7, 2017 #4


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    Science Advisor

    Right. So the current is likely much much less than 0.5A. So what does this mean for the voltage drop across the resistor and the voltage across the plates?
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