- #1
AStaunton
- 100
- 1
r=\frac{1}{2\pi\epsilon_{0}}\frac{Z_{1}Z_{2}}{mv^{2}}
The above formula is derived from coulomb's law. If we are talking about two same charges, then the force is repulsive, if two opposite charges, then attractive.
either the form of the above equation remains the same, only the meaning of the variables is slightly different, if two same charges the "r" in the equation stands for distance of closest approach and the "v" stands for initial velocity. if two opposite charges, r is the initial distance between the charges and v is the escape velocity needed (it is also the initial velocity as well I suppose).
My question is, how is it that the equation for both cases has the exact same form, since case 1 is pretty much the exact opposite of case 2, I would intuitively expect that the equations that describe them should differ by at least one minus sign, for example:
case 1:
r=\frac{1}{2\pi\epsilon_{0}}\frac{Z_{1}Z_{2}}{mv^{2}}
case 2:
r=-\frac{1}{2\pi\epsilon_{0}}\frac{Z_{1}Z_{2}}{mv^{2}}
<----i know case 2 equation is not correct, just to for illustrative purpose.
Looking at the correct equation it is clear to me that it is correct and it works and so on...
My question is what happened to the extra minus sign? is it because we can choose the zero point of potential arbitrarily for convenience and so the zero point is chosen for case 1 and case 2 so that there is no minus sign?
The above formula is derived from coulomb's law. If we are talking about two same charges, then the force is repulsive, if two opposite charges, then attractive.
either the form of the above equation remains the same, only the meaning of the variables is slightly different, if two same charges the "r" in the equation stands for distance of closest approach and the "v" stands for initial velocity. if two opposite charges, r is the initial distance between the charges and v is the escape velocity needed (it is also the initial velocity as well I suppose).
My question is, how is it that the equation for both cases has the exact same form, since case 1 is pretty much the exact opposite of case 2, I would intuitively expect that the equations that describe them should differ by at least one minus sign, for example:
case 1:
r=\frac{1}{2\pi\epsilon_{0}}\frac{Z_{1}Z_{2}}{mv^{2}}
case 2:
r=-\frac{1}{2\pi\epsilon_{0}}\frac{Z_{1}Z_{2}}{mv^{2}}
<----i know case 2 equation is not correct, just to for illustrative purpose.
Looking at the correct equation it is clear to me that it is correct and it works and so on...
My question is what happened to the extra minus sign? is it because we can choose the zero point of potential arbitrarily for convenience and so the zero point is chosen for case 1 and case 2 so that there is no minus sign?