Relationship between Electric Field and Potentential

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The discussion centers on the relationship between electric field (E) and electric potential (V), specifically the difference in their dependence on distance, with E varying as 1/r² and V as 1/r. It highlights that the electric field represents force per unit charge, while electric potential represents energy per unit charge. The connection between the two is clarified through the mathematical relationship E = -∇V, indicating that the electric field is the gradient of the potential. The inverse square relationship of the electric field arises from its derivation from the potential function. Understanding this relationship is crucial for grasping the fundamental concepts of electrostatics.
spencerthought
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E=keQ/r2 and V=keQ/r , I'm curious as to what what causes the relationship to r vs r2. I know the electric field is a measure of force per unit charge and the electric potential is a measure of energy per unit charge, just struggling to mentally connect where the inverse square relation to distance loses the square.
 
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One is the derivative of the other. The derivative of 1/r gets you 1/r^2 (save some factor).
 
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spencerthought said:
E=keQ/r2 and V=keQ/r , I'm curious as to what what causes the relationship to r vs r2. I know the electric field is a measure of force per unit charge and the electric potential is a measure of energy per unit charge, just struggling to mentally connect where the inverse square relation to distance loses the square.


The most general form of the relationship between E and V is

E = -\nabla V

This means that E is the vector gradient of the potential field V.

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