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

• instantresults
In summary: The voltage drop across the resistor is V=IR and the voltage across the plates is V=Einitial-Vdrop across resistor. So Vinitial=Einitial-Vdrop across resistor.
instantresults

Homework Statement

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

V=IR
E = V/L

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 can't 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

How can 0.5A flow in the circuit? How does the current get across the gap between the cathode and the anode?

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?

instantresults said:
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?

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?

1. How do you derive the equation E=(V-IR)/L?

The equation E=(V-IR)/L can be derived by using Ohm's Law and the equation for the electric field in a uniform electric field. First, we can substitute Ohm's Law (V=IR) into the equation for the electric field (E=V/d) to get E=(IR)/d. Then, we can rearrange this equation to get E=(V-IR)/d. Finally, we can substitute the length of the uniform electric field (d=L) to get the final equation E=(V-IR)/L.

2. What is Ohm's Law and how is it related to the electric field equation?

Ohm's Law states that the voltage (V) across a conductor is directly proportional to the current (I) flowing through it, with the proportionality constant being the resistance (R). This can be represented by the equation V=IR. This equation is related to the electric field equation because the voltage (V) in Ohm's Law is equivalent to the electric field (E) in the electric field equation (E=V/d).

3. Can you explain the significance of the length (L) in the equation E=(V-IR)/L?

The length (L) in the equation E=(V-IR)/L represents the distance between the two points where the voltage (V) and resistance (R) are measured. In other words, it represents the length of the uniform electric field. This length is important because it determines the strength of the electric field, with a larger length resulting in a weaker electric field and a smaller length resulting in a stronger electric field.

4. What is the purpose of using the equation E=(V-IR)/L in this problem?

The equation E=(V-IR)/L is used to calculate the electric field (E) in a uniform electric field, given the voltage (V), resistance (R), and length (L). It allows us to relate the voltage and resistance to the strength of the electric field and determine how these factors affect each other.

5. How can we solve this multiple choice problem using the derived equation E=(V-IR)/L?

To solve this multiple choice problem, we can plug in the given values for voltage, resistance, and length into the equation E=(V-IR)/L. This will give us the electric field (E) in units of volts per meter (V/m). We can then compare this value to the given choices and choose the answer that matches the calculated electric field. Alternatively, we can rearrange the equation to solve for a different variable, depending on what the question is asking for.

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