Electric field inside a resistor

In summary, a resistor with a resistance of 20 ohms, length of 3mm, and diameter of .1mm has a potential of 8 volts and .4 amps of current running through it. To find the electric field inside the wire, we can use the equation ΔV = - ∫ E ds, which is not based on Gauss's law but on the relationship between field and potential. In this case, the magnitude of the field is E = ΔV/L.
  • #1
SparkimusPrime
35
0
A resistor (20 ohm) is made of a very thin piece of metal wire, length = 3mm and diameter = .1mm. Given that it has a potential of 8 volts and .4 amps of current running through it, what is the electric field inside the wire?

I know there's no electrostatic equilibrium here, so I can't just take the easy answer and say there is no electric field internal to a cylinder. It seems I must use the equation that relates voltage to electric field:

V = integral(E * ds)

But, as this seems to be based upon Gauss's law (which I haven't the firmest hold) I hate to assume that I can use the equation when the charge is in motion (current).
 
Physics news on Phys.org
  • #2
Your equation is almost right: ΔV = - ∫ E ds. This is not based on Gauss's law, but on the relationship between field and potential.

In your example, the magnitude of the field is E = ΔV/L.
 
  • #3


You are correct in your thinking that the equation V = integral(E * ds) is based on Gauss's law, which states that the electric flux through a closed surface is equal to the charge enclosed by that surface. However, this equation can still be used to find the electric field inside a resistor, even when there is a current running through it.

In this case, the resistor is a thin wire with a length of 3mm and a diameter of 0.1mm. We can assume that the wire is cylindrical in shape, and the electric field will be uniform inside the wire. Using the equation V = integral(E * ds), we can solve for the electric field inside the wire by first calculating the voltage drop across the wire.

The voltage drop across the wire can be found using Ohm's law, V = IR, where V is the voltage, I is the current, and R is the resistance. In this case, the resistance of the wire is 20 ohms, so the voltage drop across the wire is 8 volts.

Next, we can plug this value into the equation V = integral(E * ds) and solve for the electric field, E. Since the wire is cylindrical, the integral becomes V = E * integral(ds), which simplifies to V = E * length. Plugging in the values we know, we get 8 volts = E * 3mm. Solving for E, we get an electric field of 2.67 volts per millimeter inside the wire.

It is important to note that this electric field is only present while there is a current running through the wire. Once the current is turned off, the wire will return to electrostatic equilibrium and there will be no electric field inside the wire.
 

1. What is an electric field inside a resistor?

The electric field inside a resistor is a vector quantity that describes the strength and direction of the electric force experienced by charged particles inside the resistor.

2. How is the electric field inside a resistor created?

The electric field inside a resistor is created by the movement of electrons through the resistor. As the electrons flow, they experience a resistance, which causes an electric field to be generated.

3. How is the electric field inside a resistor measured?

The electric field inside a resistor can be measured using a voltmeter. By measuring the voltage drop across the resistor and knowing the resistance, the electric field can be calculated using Ohm's law (E = V/R).

4. How does the electric field inside a resistor affect the flow of current?

The electric field inside a resistor acts as a resistance to the flow of current. The stronger the electric field, the more resistance there is, and the slower the flow of current will be.

5. Can the electric field inside a resistor be changed?

Yes, the electric field inside a resistor can be changed by altering the resistance of the resistor. This can be done by changing the material or dimensions of the resistor, or by adding other components such as capacitors or inductors.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
932
  • Introductory Physics Homework Help
2
Replies
44
Views
742
  • Introductory Physics Homework Help
Replies
11
Views
766
  • Introductory Physics Homework Help
Replies
17
Views
327
  • Introductory Physics Homework Help
Replies
7
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
26
Views
474
  • Introductory Physics Homework Help
Replies
1
Views
676
  • Introductory Physics Homework Help
Replies
3
Views
877
Back
Top