Explaining Positive Charge Higher Potential than Negative Charge

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A positive charge is considered to have a higher potential than a negative charge due to the nature of electrostatic potential, which is defined as the minimum work needed to move a unit positive charge from infinity to a specific point in an electric field. When moving a positive charge against the electric field (upstream), work is done, increasing potential energy. Conversely, a negative charge allows a positive charge to move with the field (downstream), decreasing potential energy. This relationship explains why a potential of +16V is greater than -18V, despite the larger magnitude of -18V. Understanding these principles clarifies the concept of potential in electrostatics.
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Homework Statement

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it is given in my high school textbook that a positive charge is that higher potential than negative charge? i am not able to understant it...can anyone please explain...

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Look at your formula for potential. What happens if you replace q with -q?
 
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Start with the definition of electrostatic potential. What is it?
 
well the definition says,"electrostatic potential at any point in a region of electrostatic field is the minimum work done in carrying a unit positive charge(without acceleration) from infinity to that point"
 
Yes. Calculate the work done in bringing a unit positive charge from infinity to that point. Do you know the electric field produced by a point charge as a function of radius?
 
calculating the work done in bringing a unit positive charge from infinity to that point in a region of electrostatic field will mean deriving formula formula for electrostatic potential due to a point charge... so you mean to say what dreamlord said that is replace q with -q in the electrostatic potential formula due to a point charge... electric field varies inversly to square of radius...
 
Yes, I am asking you to do what DimReg asked you to do in post #2, except derive it instead of using the formula.

Another way of looking it is in terms of 'upstream' and 'downstream' field lines. Consider a unit positive charge in space. When you have a charge +q at the origin, you will have to apply a force to the test charge to move it 'upstream' against the field lines. This will raise the energy of the system. On the other hand, if you have a charge -q, you can simply leave the unit positive charge and it will move 'downstream' with the field lines. This will reduce the energy of the system as the test charge will gain kinetic energy.

Basically, when there is repulsion, the potential energy increase, and when there is attraction, the energy decreases, because the system achieves a more stable state.
 
so this will mean +16v will be greater than -18v even though magnitude wise 18 is greater than 16...
 
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