I have a doubt regarding how the electric field acts in a circuit.
I have been told that a normal cell creates a uniform electric field, but I find it a bit confusing. Let me explain my doubt with a diagram.
Suppose there is a circuit and the resistor $r_1$ undergoes a potential drop of $0.4~\rm V$ and resistor $r_2$ undergoes a potential drop of $1.1~\rm V$ but then, since the electric field increases with increase in potential gradient, then the electric field through the resistor $r_2$ should be greater, shouldn't it? This goes against the idea that electric field in a circuit is uniform and if the electric field isn't constant in the circuit, then why is the battery called a fixed voltage source if the electric field applied by it varies? I find this very confusing,
Best Answer
It is best not to think in terms of fields if you are doing circuits. The primary quantities of interest in circuits are voltage and current. The E field is the spatial derivative of the voltage, eg the units of the E field is V/m. However, in a circuit the position is abstracted away. There is no indication of the length of any wire or resistor or battery. So there is no way to obtain information about the E fields given only information about the circuit theory representation of the circuit.
Inside a resistor the E field should be more or less uniform, with a value approximately equal to the voltage across it divided by its length.
Inside a capacitor the E field should be more or less uniform between the plates and zero within the plates.
Inside a battery the E field should be quite strong right at the electrode where the electrochemical reaction makes the voltage difference occur over molecular-scale distances. The E-field will be substantially lower elsewhere, including approximately 0 within the metallic conductor and relatively close to 0 within the electrolyte. Calling it uniform is not accurate.