# Tag Info

2

It seems to me that you have more of a conceptual issue than a mathematical one. To hopefully remedy this, let me remind you of a couple of facts. Given an electric field $\mathbf E$, an electric potential $V$ for $\mathbf E$ is any scalar function $V$ for which \begin{align} \mathbf E = -\nabla V \end{align} It follows that if $V$ is such a potential, ...

1

I've posted an answer describing the derivation of potential energy which you might want to read, as the same argument applies to electrical potential and I think that's what you're missing. Basically, given an electric field, the first step in finding the electrical potential is to pick a point $\vec{x}_0$ to have $V(\vec{x}_0) = 0$. Then, to determine the ...

0

In Cartesian coordinates, you have $$V(x,y,z)=E_0x,$$ so converting to spherical coordinates (using Mathematica 9.0) yields the potential $V(r,\theta,\varphi)$ of TransformedField["Cartesian" -> "Spherical", E0 x, {x, y, z} -> {r, \[Theta], \[CurlyPhi]}] $$E_0 r \sin (\theta ) \cos (\varphi ).$$ I skipped the step where you convert $\mathbf{E}$ to ...

1

Although the answer given by soliton is sufficient, I've found a way to explicitly evaluate this integral (in case anybody might be interested). Let us start from equation $(2)$ in the original message: I = \int\limits^1_0 \mathrm{d} x \int \frac{d^d p'_\text{E}}{(2 \pi)^d} \frac{1}{\left[p_\text{E}'^2 + q_\text{E}^2 x(1-x) + m^2 \right]^2} ...

0

I see the problem you have here, change the r*r drdϕdz there to z*r drdϕdz, then you should get the correct answer

4

You can prove a general formula $$\int\frac{d^dp_{\mathrm{E}}}{(2\pi)^d} \frac{(p_{\mathrm{E}}^2)^m}{(p_{\mathrm{E}}^2+\Delta)^n}= \frac{1}{(4\pi)^{d/2}}\frac{\Gamma(m+d/2)\Gamma(n-m-d/2)}{\Gamma(d/2)\Gamma(n)} \left(\frac{1}{\Delta}\right)^{n-m-d/2},\quad n>m+d/2$$ by using Gaussian integral and Euler integral of the first kind.

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