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I have some issues with the sign of the solution that I found. The problem is

Two cylindrical conductors of radii R, separated by a distance $2d$, are charged with a potential difference $V$. Find the capacitance per unit of length for the system. Suppose that $2d\gg R$.

My attempt is: $$ \phi_A-\phi_B=-\int_\infty^A\mathbf{E}(x)\cdot d\mathbf{s}+\int_\infty^B\mathbf{E}(x)\cdot d\mathbf{s}=\int_A^B\mathbf{E}(x)\cdot d\mathbf{s}=\int_A^B E(x)(-\hat{\mathbf{i}})\cdot dx(-\hat{\mathbf{i}})\\=\int_A^B E(x) dx $$ From the figure, I know that $A=d-R$ and $B=-d+R$ and by Gauss' Law the electric field in $x$ is $$\mathbf{E}=\frac{Q}{2\pi\epsilon_0 L}\left(\frac{1}{d+x}+\frac{1}{d-x}\right)(-\hat{\mathbf{i}})$$ Therefore the potential difference is $$ \phi_A-\phi_B=\frac{Q}{2\pi\epsilon_0 L}\ln\left[\frac{R}{2d-R}\right] $$

My problem is that the potential near $A$ is higher than the potential in $B$, so that the difference $\phi_A-\phi_B$ is positive, but the logarithm in the result is negative because $R<(2d-R)$.

What I am missing?

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