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two equal positive charges are kept at points A and B.the electric potential at that points between A and B (excluding these points) is studied while moving from A to B.the potential- continuously increases continuously decreases increases then decreases decreases then increases

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6 months ago

Arun
24742 Points
```							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 dd. So the potential (V) of a point at a distance rr 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 - parabola (y=kx²) 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.
```
6 months ago
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