# Charge accumulated at two ends of resistor/inductor?

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1. May 9, 2015

### kelvin490

In a simple circuit are there any charges accumulated at two ends of resistor? There is an conservative and thus electrostatic electric field and potential difference across the resistor. It seems it is only possible if positive and negative charges are accumulated at two ends to provide the electric field. Can these kind of extra charges be detected in a real resistor? How?

Same questions for inductor in the case of changing current in inductors. Thank you.

2. May 9, 2015

### Delta²

yes in both cases . To be more accurate there is surface charge density *distributed properly* along the whole surface of the resistor or the wire the inductor is made and not only at the edges. What i mean by *distributed properly* is that if for example the wire makes a curve, there has to be surface charges along the curve such as to provide the proper electric field for the electrons to follow the curved path.

As to why the charges are in the surface of the conductor (most of them) is something that follows if we apply maxwell's equations and ohm's law $\vec{J}=\sigma\vec{E}$ for the interior of any conductor with conductivity $\sigma$. For details check post #7 in this thread https://www.physicsforums.com/threads/electric-field-inside-a-closed-conductor.812046/#post-5097477

Last edited: May 9, 2015
3. May 9, 2015

### kelvin490

Thanks for the answer. I would also like to ask are these accumulated charges detectable by some devices?

4. May 9, 2015

### Delta²

Detectable yes by a simple voltmeter, but not measurable.To calculate them one should know the exact geometry of the circuit, the voltage and current sources in it, all the resistances and inductances and capacitances in the circuit, and then solve the maxwell's equations for the charge density $\rho$ of the circuit. This is very hard thing to do, can be done only with the help of computer programs.

When we connect the voltmeter in two points along a resistor, it is the surface charges at those two points that redistribute again along the leads of the voltmeter and thus create an electric field and a current inside the voltmeter and so they make the voltmeter to operate.

Last edited: May 9, 2015
5. May 10, 2015

### kelvin490

Thanks. There is a lectures explain it in terms of accumulated charges ( ). From 41:00 to 45:00 he explained that charges are build up and conservative field was set up. In 43:00 to 43:30 he said there are always some charges to keep the E field.