Question regarding potential formulas

[Carl140]

I know that the electric potential for a continuous distributions of charge can be calculated
using the following formula:

Integral [ p(y) grad( 1/(|x-y)) d A_y ]

Where grad represents the gradient of the vector function |x-y| and A_y is a area length element and p(y) represents the charge density.
I know this result is valid for 3 dimensions.

I need the result for 2 dimensions, I remember the answer involves an expression like
grad( log(|x-y|)) where log represents the natural logarithm but I can't find it anyhwere.

Anyone knows where I can find it or how to derive it? I'm pretty sure it involves a logarithm.
I've searched in Jackson and Griffiths but couldn't find it.

Thanks in advance

 

[kuruman]

Integral [ p(y) grad( 1/(|x-y)) d A_y ]
Where grad represents the gradient of the vector function |x-y| and A_y is a area length element and p(y) represents the charge density.
I know this result is valid for 3 dimensions.

This can't be right for 3 dimensions as you claim. It looks like the result for 2 dimensions, see how it's put together below.

The derivation for a two-dimensional surface is the same as a three dimensional surface: Consider the charge distribution as consisting of elements of charge $dq$ and use superposition to find the potential at $\vec r$. Using $\vec r'$ to denote the coordinate of $dq$, one writes,$$dV=\frac{1}{4\pi \epsilon_0}\frac{dq}{|\vec r-\vec r'|}=\frac{1}{4\pi \epsilon_0}\frac{\sigma(\vec r')dA'}{|\vec r-\vec r'|}$$where $\sigma(\vec r')$ is the surface charge density. The total potential is obtained by integrating primed coordinates over the surface of the two dimensional distribution,$$V(\vec r)=\frac{1}{4\pi \epsilon_0}\int \frac{\sigma(\vec r')dA'}{|\vec r-\vec r'|}.$$I am not sure where the logarithmic expressions would come in. What you remember is probably the result of integration over a specific surface.