Potential drop inside a circuit as a E=gradV

In summary, the conversation discusses the potential drop in a wire and questions why it does not occur even in the absence of a resistor. It is explained that in a real wire, there is a finite resistance which causes voltage loss, but it is usually negligible. The conversation also touches on the role of electric fields in the potential drop and how charges in a conductor can move in response to the field, resulting in a similar voltage reading as between two plates of a capacitor.
  • #1
Joker93
504
36
Considering that the electric field exist outside a battery in a wire,shouldnt the potential drop while we move along the wire even if there is no resistor(E=grad(V))?Because when i see diagrams of potential along the wire,they all show a constant potential along the wire until it reaches a resistor in which the potential drops.
 
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  • #2
You are correct. A real wire has finite resistance and you will lose voltage as you go along the wire. Usually the voltage drop in the wire is so much less than the rest of the circuit that it can be ignored, but not always.
 
  • #3
@Drakkith i am talking about something else.forget the resistance of the wire.We know from electrostatics that voltage drops gradually when there is an electric field.So,why does the potential not drop inside the circuit when we move along the wire?it has nothing to do with resistance but everything to do with electric fields
 
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  • #4
Because a conductor is composed of charges that can move in response to the electric field. If you attach a long wire with a small capacitance to each plate of a capacitor, the charges in the wires move in response to this electric field. Measuring the voltage between the ends of these wires will give you approximately the same voltage as there is between the two plates because the charges have moved in response to the field and accumulated, setting up their own electric field that gives you a voltage between the ends of the wires.
 

Related to Potential drop inside a circuit as a E=gradV

1. What is the concept of potential drop in a circuit?

The concept of potential drop in a circuit refers to the decrease in electrical potential energy as an electric charge moves through a circuit. This is caused by the resistance of the circuit components and can be calculated using the formula E=gradV, where E is the potential drop and gradV is the change in voltage.

2. How is potential drop measured in a circuit?

Potential drop is measured in volts (V) using a voltmeter. The voltmeter is connected in parallel to the circuit components and measures the difference in potential between two points in the circuit, showing the potential drop.

3. What factors affect the potential drop in a circuit?

The potential drop in a circuit is affected by the resistance of the circuit components, the current flowing through the circuit, and the length of the circuit. The higher the resistance, the greater the potential drop. Similarly, a higher current and longer circuit length will also result in a larger potential drop.

4. How does potential drop affect the efficiency of a circuit?

Potential drop can have a significant impact on the efficiency of a circuit. The higher the potential drop, the more energy is lost as heat through the resistance of the circuit components. This can result in a decrease in the overall efficiency of the circuit, as less energy is being used for its intended purpose.

5. Is potential drop the same as voltage?

While potential drop and voltage are related concepts, they are not exactly the same. Voltage refers to the potential difference between two points in a circuit, while potential drop specifically refers to the decrease in voltage as an electric charge moves through a circuit. The potential drop can be calculated using the voltage difference and the resistance of the circuit components.

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