I get this, but I don't understand why assume that the potential in đ” is lower than in đŽ. The direction of the electric field lines is the direction of the force a positive charge would experience if placed in the field. In order to move a positive charge from B to A which is against the field external work must be done on the charge. This increases its potential energy. If the positive charge is placed at A it will naturally move from A to B, and therefore loses electrical potential energy. The analogy with a gravitational field and gravitational potential energy may be helpful. In the figure below I have rotated your diagram. Imagine now that instead of the lines being an electric field, that they represent a uniform gravitational field. Like our positive electric charge, the gravitational field lines point in the direct of the force that an object would experience if placed in the gravitational field. If you raise an object from B to A, against the gravitational field, you know you have to do work against the gravitational force and that the object gains gravitational potential energy. If you release an object from rest at A it falls losing potential energy and gaining kinetic energy. The point B is said to be at a lower gravitational potential than A. It's the same case for electrical potential. The point B is at a lower electrical potential then the point A. Hope this helps. 
Electric field lines always travel from higher potential to lower potential So VA is greater than VB so difference between VB and VA us negative
VBââVAâ=(âVeâ) because, It goes to electric field always goes higher potential to lower potential. but, VBâ to VAâ is the opposite direction.. So potential difference VB-VA= negative |
Two points A and B are located within the electric field
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