Can a charge in vacuum produce uniform electric field on its own? I know that uniform electric fields are having evenly spaced field lines and the intensity of the field at every point must be the same. 
When I confronted my teacher with the question above she said that a charge can produce a uniform external field over a finite distance, but I feel unconvinced.My textbook even states that the electric field produced by a charged particle varies with position , if it's so then how can a charged particle , with nothing to disturb it produce a uniform external field around it?
I have not much information about how uniform electric field can be produced, someone told me that they exist between charged plates.
 A: Let me put together some of the things mentioned in the comments. For simplicity, let's consider the case of sources and fields in vacuum. Also, I'll work in 2D but the same result follows in 3D or 1D.
What can one do to obtain a uniform or approximately uniform electric field? One simple alternative is to consider a localized charge distribution, exploring the field far enough from it and/or focusing on a charge-less region small enough (without abandoning classical physics). This would be an approximately uniform field even for the radial $1/r^2$ field of a single charge. We can quantify how close this field is to being uniform. In the process we will find that there are more effective ways of producing an approximately uniform field.
Let's place a singe charge in the origin. Let's say we want to see how the field changes in a region at a distance $R$ to the right. The field in the surrounding area with this shift of the coordinates is
\begin{equation}
\mathbf{E}_{\mathrm{Q}} = \frac{Q}{4 \pi \epsilon_0} \frac{1}{\left[\left(R+x\right)^2+y^2\right]^{3/2}}\left[\left(R+x\right) \hat{\mathbf{i}} + y \,\hat{\mathbf{j}} \right] \, ,
\end{equation}
where we can see $x$ and $y$ as the coordinates relative to this far away point. Now we can expand for large $R$ and find
\begin{equation}
\mathbf{E}_{\mathrm{Q}} = \frac{Q}{4 \pi \epsilon_0} \left[\left(\frac{1}{R^2}- \frac{2 x}{R^3}\right)\hat{\mathbf{i}} + \frac{y}{R^3} \,\hat{\mathbf{j}} \right] \, + \, \mathcal{O}\left(R^{-4}\right) .
\end{equation}
Hence, the dominant deviation from a uniform field goes as $\frac{x}{R^3}$ or $\frac{y}{R^3}$. But we can do better.
Consider now a dipole, more precisely, a positive charge $Q$ placed at $(-R,0)$ and a negative charge $-Q$ at $(R,0)$. As we mentioned, if we take $R$ big relative to the area of focus, we obtain an approximately uniform field near the origin. The exact field is given by
\begin{eqnarray}
\mathbf{E}_{\mathrm{d}} &=& - \mathbf{\nabla} \Phi \,\, \quad \, \, \mathrm{with}
\\
\Phi &=& \frac{Q}{4 \pi \epsilon_0} \left[ \frac{1}{\sqrt{\left(x+R\right)^2+y^2}} - \frac{1}{\sqrt{\left(x-R\right)^2+y^2}} \right] \, .
\end{eqnarray}
We can  now play the same game and consider the limit of large $R$ which leads us to
\begin{equation}
\mathbf{E}_{\mathrm{d}} = \frac{Q}{2 \pi \epsilon_0} \left[ \left( \frac{1}{R^2}+ \frac{6 x^2-3y^2 }{2 R^4}  \right) \hat{\mathbf{i}} - \frac{3 x y}{R^4} \, \hat{\mathbf{j}} \right] \, + \, \mathcal{O}\left(R^{-6}\right) .
\end{equation}
In this case, the dominant deviations from a uniform field go as $\frac{x^2}{R^4}, \frac{y^2}{R^4}$ or $\frac{xy}{R^4}$. Consequently, this field converges faster to a uniform field for $x,y \ll R$. Plotting shows this nicely.
Finally, we can use this final result to show how to obtain an exact uniform field from the dipole field. This is done by first taking
\begin{equation}
Q = 2 \pi \epsilon_0 R^2 E_0 \, ,
\end{equation}
with $E_0$ a constant. This is, in other words, keeping the fraction $\frac{Q}{R^2}$ constant. Then one takes the limit $R \rightarrow \infty$ leading to $\mathbf{E}_{\mathrm{d}} \rightarrow E_0 \hat{\mathbf{i}} \, $. While it might seem like an arbitrary trick, this limit has real physical applications, like when one uses the method of images to compute the electric field of a conducting spherical shell in the presence of a uniform electric field. One starts with a "finite-distance" dipole and locates the corresponding image charges that keep the surface of the sphere as an equipotential. Then one takes the limit we discussed to find the solution. For a discussion of this problem, see section 2.5 of Jackson's book, Classical Electrodynamics, 3ed.
A: An isolated charged point particle cannot produce a homogeneous electric field. It has a radial $1/r^2$ dependence and is thus spherically symmetric. If you have many charged particles (electron) on a metal plate, you can to a very good approximation assume that the charge is continuously distributed over a large center part of the plate so that you will get a homogeneous electric field normal to the surface there. If you have an infinitely extended surface charge, the electrical field will be perfectly homogeneous.
Added note: In a metal you have an enormous density of electrons $n ≈ 10^{22}cm^{-3}$  Thus you can consider the electrons in a metal as point charges producing a statistically virtually continuous charge density $\rho (\vec r')=-e\cdot n(\vec r')$ which gives you a practically continuous Coulomb electric field $$\vec E (\vec r)=\frac {\rho (\vec r')(\vec r-\vec r')}{4\pi \epsilon |\vec r-\vec r'|^3}$$ You could also consider the superposition of the individual electrons Coulomb fields at any point outside with the same result. The situation becomes even more smoothed out when you consider the electrons quantum-mechanically. The electrons in a metal are not localized and the charge of each individual electron can be considered to be smeared out over a large region of the metal.  
