A rod of length $L$ lies along the x axis with its left end at the origin. It has a non uniform charge density $\lambda = \alpha x$ where $\alpha$ is a positive constant. Calculate the electric potential at $A(-d,0,0)$.

First Attempt

$$V = - \int \vec{E}\cdot \mathrm d\vec{l}$$

Calculating The electric field $E$:

$$E = \int \frac{\lambda\,\mathrm dl}{4\pi \epsilon r^2 }$$

with $r^2 = (x+d)^2$

We then get $$E=\frac{\alpha}{4 \pi \epsilon}\Big(\ln(x+d) + \frac{d}{x+d}\Big)$$

Then, $$V = - \int_0^L E\cdot\mathrm dl= -(L+2d)\ln(l+d)+l+2d\ln(d)$$

Second Attempt

Using the formula $$V = \int_0^L \frac{\lambda\,\mathrm dl}{4\pi \epsilon r}$$

Evaluating this integral leads to the following :

$$V=\frac{\alpha}{4 \pi \epsilon}\Bigg(L-d \ln\Big(\frac{L+d}{d}\Big)\Bigg)$$

Why do both attempts give different solutions?


closed as off-topic by ZeroTheHero, Jon Custer, Kyle Kanos, Aaron Stevens, Qmechanic Jan 30 at 16:32

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  • 2
    $\begingroup$ There are many issues here. In the first method you shouldn't have $r^2=(x+d)^2$. If you wan to go the first attempt route, you are trying to find the potential difference between the point A and the left end of the rod. So you need to find the field at all points on the line from $x=0$ to $x=-d$ (so maybe say $r^2=(x+x')^2$ and do the first integral over $x$?), and then do the line integral over $x'$ from $x=0$ to $x=-d$ $\endgroup$ – Aaron Stevens Jan 30 at 15:36
  • 2
    $\begingroup$ Another issue is that the first method puts $V=0$ at $x=0$, but the second method puts $V=0$ at infinity. I would just stick with the second method. $\endgroup$ – Aaron Stevens Jan 30 at 15:37