Potential energy is zero at infinite distance We set the electrical potential energy to be zero in infinite distance to calculate potential energy at a distance $r$. But how is it possible from infinite distance to end up with finite distance? How we measure the work done by electric force? 
 A: It's a matter of convention setting the potential of inverse square fields to zero at infinity because the fields themselves fall as $1/r^{2}$. For electrostatic fields, the work done is 
$$ W=\int_{r_{i}}^{r_{f}}F(r)dr =e\int_{r_{i}}^{r_{f}}E(r)dr$$
when you wish to move an electric charge from position $r_{i}$ to $r_{f}$ in the presence of an electric field. The term in the second integral is defined as the electric potential. In general vector notation it is
$$ V(\mathbf{r})=-\int_{\mathcal{C}}\mathbf{E}(\mathbf{r}')\cdot d\mathbf{r}' $$
The point is that, if you want to talk about the electric potential about a particular point, it has to be in reference to another suitably defined one. Hence we used a point at infinity for this purpose, because we know that any influence of electric fields would vanish.
A: When opposite charges are far apart the potential energy is infinite but the force is small.  When the opposite charges are together there is no more potential energy.  But it is the opposite for similar charges.  Work is force times distance, or the area under a curve of a plot of force versus distance.
A: Work done by a force  is defined by $$W=\int^{\vec{r_1}}_{\vec{r_2}}\vec{F} \cdot d\vec{s}$$
where $\vec{r_1}$ and $\vec{r_2}$ are the position vectors of the start and end points.
Are you worried that when you evaluate this integral from the infinity point to a finite point, the result will be infinity? The easiest way to check this is to perform the integral!
Suppose a charge $+Q$ is resting at the origin $(0,0,0)$ of a 3D Cartesian coordinate system. The electric field $\vec{F}_e$ along the $x$ axis will be given by
$$\vec{F}_e=\frac{1}{4\pi\epsilon_0}\frac{Q}{x^2}\hat{\vec{x}},$$
where $\hat{\vec{x}}$ is the unit vector in the positive $x$ direction.
Next suppose a charge $+q$ is resting on the $x$ axis, at the point $(\infty,0,0)$.
We want to bring this charge from the infinity point $(\infty,0,0)$ to a finite point $(A,0,0)$ and calculate the work done by the electric force acting on charge $+q$.
The force $\vec{F}$  acting on charge $+q$ along the $x$ axis is given by $$\vec{F}=\frac{1}{4\pi\epsilon_0}\frac{Qq}{x^2}\hat{\vec{x}},$$
where $x$ is the coordinate of charge $+q$.
Thus the integral we want to calculate is
$$W=\int^{\vec{r_1}}_{\vec{r_2}}\vec{F} \cdot d\vec{s}$$
$$=\int^{\vec{r_1}=A\hat{\vec{x}}}_{\vec{r_2}=\infty\hat{\vec{x}}}\frac{1}{4\pi\epsilon_0}\frac{Qq}{x^2}\hat{\vec{x}}\cdot d\hat{\vec{x}}$$
$$=\int^{x=A}_{x=\infty}\frac{1}{4\pi\epsilon_0}\frac{Qq}{x^2}dx$$
$$=\frac{Qq}{4\pi\epsilon_0}\int^{x=A}_{x=\infty}\frac{1}{x^2}dx$$
$$=\frac{Qq}{4\pi\epsilon_0}[-\frac{1}{x}]^A_\infty  $$
$$=\frac{Qq}{4\pi\epsilon_0}[-\frac{1}{A} + \frac{1}{\infty}]$$
$$\approx-\frac{Qq}{4\pi\epsilon_0}\frac{1}{A}$$
This is the expression for the work done by the electric force acting on charge $+q$.
So, even if the charge $+q$ travels an infinite distance, we still get a finite expression for work done!
