- #1
indigojoker
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A thin wire has charge +Q uniformly distrubuted along it's length L. It's lying on the x-axis with its midpoint at the origin. Calculate the potential at a point P on the x axis, where P>L/2.
[tex] dV= \frac{k dq}{r} [/tex]
[tex]V = k \lambda \int \frac{1}{r}dr[/tex]
[tex]V = k \lambda \int^{P+L/2} _{P-L/2} \frac{1}{r}dr[/tex]
So when i use that integral, i get:
[tex] V = k \lambda \ln {\left( \frac{P-\frac{L}{2}}{P+\frac{L}{2}} \right)} [/tex]
If i wanted to find the E field, I know that [tex] E = - \nabla V[/tex]
but do I take the derivative w.r.t. P?
[tex] dV= \frac{k dq}{r} [/tex]
[tex]V = k \lambda \int \frac{1}{r}dr[/tex]
[tex]V = k \lambda \int^{P+L/2} _{P-L/2} \frac{1}{r}dr[/tex]
So when i use that integral, i get:
[tex] V = k \lambda \ln {\left( \frac{P-\frac{L}{2}}{P+\frac{L}{2}} \right)} [/tex]
If i wanted to find the E field, I know that [tex] E = - \nabla V[/tex]
but do I take the derivative w.r.t. P?