What is 1 Volt? Class 10. Physics.
What do we mean by saying 1 volt, 100 volts, 220 volts et cetera?
Let first define 1 volt.
According to Charles Augustin de Coulomb's law or Coulomb's law two electrically charged particles interact each other by exerting a mechanical force between them. This force may either be attractive or repulsive in nature. If both the particles possess charges of same type, then the force is repulsive. And for opposite charges it is attractive.
Suppose, a huge electrical charge, containing a total electrical charge of q coulomb, is placed here. Then, theoretically it can exert force to any charged particle, wherever the particle be. But practically, intensity of its electric field is restricted within a few meters.
Now, if you have a point particle, carrying an electrical charge of plus 1 coulomb, far away from the huge charge particle. And, if you wish to place it at a nearby point of the huge charge. Suppose at that particular red point. Then, what you need to do first?
To answer this question, consider the conditions.
Well, two situations may arise. Because, the charge Q can either be a positive one, that is plus q coulomb, or a negative, or minus q coulomb. In the First case, let the huge central charge of Q be positive.
And there after, we shall discuss the second occasion, when the central charge is a negative.
Let see the first situation.
Since, both the objects carrying a positive charge. Therefore, a repulsive force is acting between them. You have to apply a certain amount of energy to place the particle to a desired position.
Why this application of energy is necessary?
Because, the force acting on the 1 coulomb particle is Outward. .
As you pushing the point charge, against the Outward repulsive force, you are actually performing a mechanical work.
Coulomb's force is directed radially outwards. And displacement is radially inwards. Hence, the force and the displacement are along a straight line. Thus the work done is force into displacement.
But, there is problem in calculating the work done. The coulomb's force is not uniform at all places. Its value depends on the distance from it. The force is higher at a closer distance. And zero at infinity. These green buildings may give you a very good approximation. Near the charge, a higher building indicates a higher value of coulomb's force. As you start going away, you could see low-height buildings, indicating a low magnitude coulomb's force.
To simplify the calculation,
Let us divide the whole path of displacement into some segments. They are of lengths, r1, r2, r3, r4, r5, r6, et cetera. We have made a total 15 numbers of segments. A shorter segment can bring an accuracy in calculation, but at a cost of a huge number of segments.
If the average value of the repulsive force throughout the path r1 is f1, and throughout the path r2 is f2 ,and throughout the path r3, is f3, and so on. Then work done on the path r1 is say w1, which is equal to minus f1 into r1.
Minus sign arises for the fact that the force and the displacement are in opposite direction.
Thus, w1 equals minus f1r1.
Similarly for the path r2, work done is equal to minus f2r2.
For the path r3, w3 equals minus f3r3.
And so on.
Summing up all these small works we get the total work for this displacement to be w1 plus w2 plus w3 plus w4 plus,, et cetera et cetera.
Electrical potential phi is defined to be equal to minus the work done.
Thus phi equals minus bracket w1 plus w2 plus w3 plus w4 plus and so on.
Then we get phi equals minus bracket minus f1r1 minus f2r2 minus f3r3 and so on.
Or phi equals plus f1r1 plus f2r2 plus f3r3 and so on.
A total of all your mechanical works are called the electrical potential at that red point.
Thus, if your total work done is 1 joule, then the electrical potential at that red point is plus 1 volt.
For a mechanical work of 100 joule, the potential is plus 100 volts. And so on......
Now let take the second situation.
The charges are opposite now. Since opposite charges attract each other. They too will do so.
If you place the plus 1 coulomb point particle at rest at a point far away from the huge charge, the particle will start getting attracted towards the huge charge. And after some certain period of time, it will reach the desired location.
Thus, in this case, you need not to apply any force on it. The system itself will have done that work for you.
The work is positive here. Because, the force and the displacement are in the same direction.
So the total work is w1 plus w2 plus w3 plus w4 plus and so on. Or w equals f1r1 plus f2r2 plus f3r3 plus and so on.
Therefore, electrical potential at that point is minus bracket f1r1 plus f2r2 plus f3r3 plus f4r4 plus and so on.
Or, phi equals minus f1r1 minus f2r2 minus f3r3 minus And so on.
Note that, in this second occasion, the electrical potential is negative.
If the work done by