Electric Intensity: Definition & Clarification

In summary: I is the number of coulombs of charge that pass through the disk per second...but I don't see where it says it has to be in 1 sec?What you are asking is not answerable.
  • #1
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It is written in my book that the Electric intensity is the amount of electric charge which passes in one second. I think the definition is incomplete, I mean it didn't mention the sector that it should pass in one second or the distance, does it mean the amount of electric charge which passes a certain point or the whole wire in one second? Please I need help in clarifying the concept of electric intensity. Thanks in advance.
 
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  • #2
This is NOT electric Intensity, this is called Electric CURRENT. And it is not necessary to mention any particular point, because there will be same current at any cross-section in the wire (if there is no branch wire).
The definition is , the amount of charges flowing per unit time through any cross-section of a wire, is called Electric Current.
Electric Intensity is a totally different topic. In an electric field if a unit positive charge is brought, then the amount of force experienced by the unit charge is called the Electric Intensity of that field at that point.
 
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  • #3
AbhiFromXtraZ said:
This is NOT electric Intensity, this is called Electric CURRENT. And it is not necessary to mention any particular point, because there will be same current at any point in the wire (if there is no branch wire).
Electric Intensity is a totally different topic. In an electric field if a unit positive charge is brought, then the amount of force experienced by the unit charge is called the Electric Intensity of that field at that point.
I mean when we say that 5 coulombs pass a certain point in the wire, what is point by point, how much does this point occupy the wire??
 
  • #4
I edited my answer...it's not point...it's cross-section of the wire..
 
  • #5
AbhiFromXtraZ said:
I edited my answer...it's not point...it's cross-section of the wire..
How much would it take from the volume of the wire ??
 
  • #6
Nothing. A cross-section is infinitely thin. For a cylindrical wire, imagine an infinitely thin circular disk that spans across the width of the wire. The current I is the number of coulombs of charge that pass through the disk per second.
 
  • #7
AbhiFromXtraZ said:
In an electric field if a unit positive charge is brought, then the amount of force experienced by the unit charge is called the Electric Intensity of that field at that point.

In English, this is usually called simply the "electric field" nowadays. In 25 years of teaching physics, and 10 years as a physics student before that, I don't remember seeing "electric intensity" in any of the textbooks that I've used.
 
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  • #8
jtbell said:
Nothing. A cross-section is infinitely thin. For a cylindrical wire, imagine an infinitely thin circular disk that spans across the width of the wire. The current I is the number of coulombs of charge that pass through the disk per second.
Thanks a lot I have been confused for a while because I couldn't find the specific description for cross section. Thanks to you, now I understand it. But this makes a little confusion about the electric power equation where it is known to be equal to VI, if we were to have a resistor of 5 ohms connected to a power source of 10 volts, obviously, I = 2 amperes so the power according to VI would be 20 watts, what I'm confused about is that Power means the amount of energy consumed per second VI = 10*2 means that a works of 20 joule was done to transfer a charge of 2 coulombs in one second, but this 20 joule should be consumed when the 2 coulombs completely pass the 5 ohm resistor which isn't necessary to be in 1 sec I mean power should be equal to V times number of coulombs which pass the 5 ohm resistor / sec, not VI as this V might not be completely consumed by these coulombs in one sec
 
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  • #9
The definition of power states how much work is done in 1 second.
In this case 10Joule is the amount of work required to pass 1 coulomb charge through 5 ohm. So 2x10J energy is requred for 2 coulomb.
So this is the work done for 2 coulomb to completely pass. But since I is 2 A, 2 coulomb charge passes in 1 second. So 2x10J is the amount of work done in 1 second.
Hence the power is 2x10W=20W.
Hope you understood.
 
  • #10
AbhiFromXtraZ said:
The definition of power states how much work is done in 1 second.

In this case 10Joule is the amount of work required to pass 1 coulomb charge through 5 ohm. So 2x10J energy is requred for 2 coulomb.

So this is the work done for 2 coulomb to completely pass. But since I is 2 A, 2 coulomb charge passes in 1 second. So 2x10J is the amount of work done in 1 second.

Hence the power is 2x10W=20W.

Hope you understood.
How can I say that in that second 10 joules were consumed, maybe the 2 coulombs wouldn't pass the 5 ohm resistor in 1 sec
 
  • #11
Remember what is I and how did you get it. I=V/R. And the voltage and resistance are constant. So from definition 2 coulomb actually passes in 1 sec. Imagine the cross section just at the end of the resistance and remember the definition of I.
And energy consumed is 20J not 10J in a second.
 
  • #12
Suppose we have a wire of uniform cross-section, carrying a current I. As explained earlier, we can define the current as the charge passing through any cross-section per unit time.

But the only reason charge passes through the cross-section is that charge-carriers (e.g. free electrons) throughout the wire have a certain mean drift velocity along the wire.

Thus throughout the wire these charge carriers are losing the extra energy (which they have because of their drift velocity) in collisions with the ion-cores in the metal (which is thereby heated). Their energy is continuously replenished by work done on them by an imposed electric field, if there is a battery connected across the wire. [When the battery is disconnected the charge carriers lose their drift velocity in a very short time indeed.]
 
  • #13
Philip Wood said:
Suppose we have a wire of uniform cross-section, carrying a current I. As explained earlier, we can define the current as the charge passing through any cross-section per unit time.
But the only reason charge passes through the cross-section is that charge-carriers (e.g. free electrons) throughout the wire have a certain mean drift velocity along the wire.
Thus throughout the wire these charge carriers are losing the extra energy (which they have because of their drift velocity) in collisions with the ion-cores in the metal (which is thereby heated). Their energy is continuously replenished by work done on them by an imposed electric field, if there is a battery connected across the wire. [When the battery is disconnected the charge carriers lose their drift velocity in a very short time indeed.]
Thanks a lot, could you explain how is electric power calculated ?
 
  • #14
Power = Current in wire x p.d. across wire = IV.
 

1. What is electric intensity?

Electric intensity is a measure of the strength of an electric field at a particular point in space. It is also known as electric field strength or electric field intensity.

2. How is electric intensity measured?

Electric intensity is typically measured in units of volts per meter (V/m) or newtons per coulomb (N/C). It can be measured using an instrument called an electric field meter.

3. What factors affect electric intensity?

The electric intensity at a point is affected by the magnitude and direction of the electric field, as well as the distance from the source of the electric field. The type and properties of the materials in the electric field can also affect the electric intensity.

4. What is the difference between electric intensity and electric potential?

Electric intensity is a vector quantity that describes the strength and direction of an electric field. Electric potential, on the other hand, is a scalar quantity that describes the potential energy of a charged particle in an electric field. In simpler terms, electric intensity tells us how strong the electric field is, while electric potential tells us how much work is required to move a charged particle in the electric field.

5. How is electric intensity used in real-world applications?

Electric intensity is a crucial concept in many areas of science and technology. It is used in the design of electrical circuits, the operation of electronic devices, and in understanding the behavior of charged particles in nature. It is also used in medical applications such as electrocardiograms (ECGs) and electroencephalograms (EEGs).

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