Electrode Voltage Difference

[levi2613]

Electrode Voltage Difference

Homework Statement

A vacuum photodiode is constructed by sealing two electrodes in a vacuum tube. The electrodes are separated by L and connected to a battery and a resistor (R). The potential difference between the cathode and anode is approximately equal to the battery voltage V. The electric field at all points between the electrodes is equal to the electrode voltage difference divided by L. Which change to the circuit will decrease the electric field by the greatest amount: increasing L by a factor of 2, or increasing the circuit resistance by a factor of 2?

Homework Equations

The solution says "For a fixed voltage between cathode and anode, the electric field is inversely proportionalto the distance between them. Increasing the circuit resistance for a fixed current will decrease the electric field, but not by as much as does the length change. E = (V-IR)/L


3. Question
I don't understand where the final equation comes from. I would think that Ohm's Law would dictate that E would then always be zero. Obviously, that's not the case, but I don't understand why, and I can't find that equation anywhere else...

Thanks so much.

 

[tiny-tim]

Kirchoff's rules

The potential difference between the cathode and anode is approximately equal to the battery voltage V.
The electric field at all points between the electrodes is equal to the electrode voltage difference divided by L.
…
E = (V-IR)/L
…
I don't understand where the final equation comes from. I would think that Ohm's Law would dictate that E would then always be zero.

Hi levi2613!

This is a circuit with three items: a battery, the cathode/anode, and the resistor.

Hint: from Kirchoff's rules applied to this circuit, the potential difference between the cathode and anode must be … ?

And then divide by L to get E, as told.

 

[Moetasim]

The equation E = (V-IR)/L comes from the formula for electric field, which is E = V/d, where V is the voltage and d is the distance between the electrodes. In this case, we can substitute d with L, and since the voltage is not constant due to the presence of the resistor (R), we use Ohm's Law (V=IR) to express V in terms of I and R. Therefore, the equation becomes E = (V-IR)/L.

This equation shows that the electric field is not directly proportional to the voltage difference (V) or the distance between the electrodes (L), but rather it is affected by the current (I) and the resistance (R) in the circuit. Therefore, changing either the distance between the electrodes (L) or the circuit resistance (R) will affect the electric field, but the change in L will have a greater impact on the electric field compared to the change in R.

It is important to note that Ohm's Law only applies to certain types of materials and under certain conditions. In a vacuum photodiode, the vacuum between the electrodes acts as an insulator, so Ohm's Law may not directly apply. However, the equation provided is a simplified representation of the electric field and it takes into account the effect of the circuit components on the electric field.