How to Decrease the Electric Field in a Vacuum Photodiode Circuit?

[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.