The electric potential:

$\overline{){\mathbf{V}}{\mathbf{=}}\frac{\mathbf{k}\mathbf{q}}{\mathbf{r}}}$

The charge carried by the piece of wire is given by,

$\overline{){\mathbf{d}}{\mathbf{q}}{\mathbf{=}}{\mathbf{\lambda}}{\mathbf{d}}{\mathbf{x}}}$, where λ is the linear charge density.

To calculate the electric potential at the center of the semicircle due to the right wire:

dV_{right} = kdq/x

Integrating both sides from x = R to 3R:

$\begin{array}{rcl}{\mathbf{V}}_{\mathbf{r}\mathbf{i}\mathbf{g}\mathbf{h}\mathbf{t}}& \mathbf{=}& {\mathbf{\int}}_{\mathbf{R}}^{\mathbf{3}\mathbf{R}}\mathbf{\left(}\frac{\mathbf{k}\mathbf{d}\mathbf{q}}{\mathbf{x}}\mathbf{\right)}\\ & \mathbf{=}& {\mathbf{\int}}_{\mathbf{R}}^{\mathbf{3}\mathbf{R}}\mathbf{\left(}\frac{\mathbf{k}\mathbf{\lambda}\mathbf{d}\mathbf{x}}{\mathbf{x}}\mathbf{\right)}\\ & \mathbf{=}& \mathbf{k}\mathbf{\lambda}{\mathbf{\left[}\mathbf{l}\mathbf{n}\mathbf{x}\mathbf{\right]}}_{\mathbf{R}}^{\mathbf{3}\mathbf{R}}\end{array}$

What is the electric potential at the center of the semicircle?

**Give your answer in terms of** *λ***,** *R***,** *ϵ *0**and appropriate constant**

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