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## Main Question or Discussion Point

the potential difference between

V(

the integral is taken from a to b.

so the potential of a positive charge, with infinity as reference, is

V(

the integral is from infinity to r.

My question is this.

Clearly, the potential of the positive charge is positive. But that means the scalar product in this integral is also positive, even though

**b**and**a**is defined as follows:V(

**b**) - V(**a**) = -∫**E**[itex]\bullet[/itex]d**l**the integral is taken from a to b.

so the potential of a positive charge, with infinity as reference, is

V(

**r**) - V(infinity) = V(**r**) = -∫**E**[itex]\bullet[/itex]d**l**the integral is from infinity to r.

My question is this.

**E**points outward, and since integral is from infinity to**r**, we would be moving inward over the integral. So d**l**, and hence the scalar product in the integral, should be negative. But since**E**is inverse square in r, integrating gives another negative sign, giving us 3 in all.This would lead to a negative potential for the positive charge.Clearly, the potential of the positive charge is positive. But that means the scalar product in this integral is also positive, even though

**E**and d**l**seem to be pointing is opposite directions.
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