equal and opposite charges Q are placed at a point A and B as shown in figure andP1 and P2 equidistant from then ratio of potential at P1 and P2
Answers
Answer:
We can look at this question from two perspectives. One in a mathematical sense and one in an intuitive and logical sense. The mathematical calculations are built on the basic axioms but it's important to understand why it follows logically. Once we have done that we can also confirm it through the math.
Logical Approach
First let's start by defining potential. It's important to know what it means.
Potential energy of a charge kept at a point is the negative of the work done on a body to bring it slowly from infinity to that point.
So basically if you have a place with many charges and stuff, the body will experience some force while you are bringing it from infinity. So you will have to do apply some force and thus do some work to make sure it moves slowly. The negative of the work done by you is the potential energy of the charge at the point.
The potential of an unit charge kept at a point is the potential of that point.
Now imagine that you have kept a positive unit test charge Q midway between A and B. The two charges kept at A and B are both equal and equidistant from Q. So both the charges pull on Q equally. So we can expect Q to not move at all.
Now we know that the force on between two charges is inversely proportional to the square of the distance between them. So if we have two charges placed closer to each other, they will pull/push each other with a stronger force than if they were farther away.
Now imagine that you place it closer to the point A.
The test charge and the charges at A and B are all positive. So we would expect them to repel each other. So the charges Q and A will push each other away harder than the charges Q and B would. So the charge Q will experience a stronger force away from A and so will have a tendency to move towards B.
Now let's say the points A and B are fixed and Q is allowed to move. So we should expect to see the charge Q accelerate towards B until it reaches the midpoint. At the midpoint, it's going to be in equilibrium, i.e. no force acts on it. However it is already moving towards B with some speed. So it just continues on towards B. However when it crosses the mid point, it is now closer to B. So now B pushes it harder. So it slows down till it stops and then starts accelerating towards A. Then it again crosses the midpoint and the same thing happens again.
One of the basic laws of nature states that any body wants to move from higher energy to lower energy.
We have seen that the unit charge wants to move from any point between A and B towards its midpoint. And at the midpoint if it doesn't have a velocity, it won't move anywhere. As everything wants to lower its energy, we can say that the surroundings points are at a higher energy while the midpoint has the lowest energy. So we can imagine that the midpoint is like a valley between two hills, like a sink.
So it logically makes sense that the potential (the potential energy of an unit charge) decreases as we move from A to the midpoint and then increases as we move from the midpoint to B.
Mathematical Approach
The potential due to a charge Q is kQ/r , where r is the distance of the point from the charge and k is the constant 1/(4πε) .
Let the distance between A and B be d . So the potential (V) of a point at a distance r along the line joining A and B is (kq /r)+(kq/(d−r)) , where q is the charge of A and B.
Therefore,
V(r) =kq(1/r+1/(d−r))=kqd/(r.(d−r))
The graph of V with r is -
Where the potential is along the y axis and the distance r is along the x axis.
So we can see that the potential decreases and then increases on moving from A to B.