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I'm trying to find the electric field from the potential and vise-versa but I'm having trouble

I know the electric field of a ring with radius a and charge Q to be $$E=\frac{Qx}{4\pi \varepsilon_{0}(x^{2}+a^{2})^{3/2} }$$

and the potential to be $$V=\frac{Q}{4\pi\varepsilon_{0}\sqrt{x^{2}+a^{2}}}$$

How can I reach the equation for potential from the electric field given that $$V=\int E\cdot dl=\int E\cos \phi dl$$

$$\cos \phi =\frac{x}{\sqrt{x^{2}+a^{2}}}$$

It seems that just solving the integral would yield the answer but somehow it is giving me completely different answer

Edit

This is the work done by me, although I am not entirely sure if this is the way to do it-

$$V=\int \frac{Qx}{4\pi \varepsilon_{0}(x^{2}+a^{2})^{3/2} }\cdot dx$$ $$=\frac{Q}{4\pi \varepsilon_{0}}\int \frac{x}{(x^{2}+a^{2})^{3/2}}dx$$ $$=-\frac{Q}{4\pi\varepsilon_{0}\sqrt{x^{2}+a^{2}}}$$

I have no idea what to do with the negative sign.

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You have made two elementary errors. First, you forgot a negative sign and second, you forgot your limits (V at infinity is conventionally taken to be zero)

$V = -\int_{\infty}^{r} E dr$

Since I believe you are taking the electric field at the x axis, instead of r you can use x, making that part of your work correct.

These two things will fix your answer.

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