Won't the test charge disturb the electric field to be measured? As written in Principles of Physics by Halliday, Resnick and Walker:

The electric field is a vector field; it consists of a distribution of vectors,one for each point in the region around a charged object. In principle, we can define the electric field at some point near the charged object. We first place a positive charge $q_0$, called a test charge,at the point. We then measure electrostatic force $\vec{F}$ that acts on the test charge. Finally, we define the electric field $\vec{E}$ at that point due to the charged object as $$\vec{E} = \frac{\vec{F}}{q_0}$$ .

So, this is the way they derive the electric field. But one thing that perplexes me is that as test charge has its own field, so when the test charge is introduced in the field of concerned charged object, say another point charge $+q$ , won't it get shifted due to the repulsion from the test charge $q_0$ (which is, by convention, positive)? Then consequently the field of $+q$ will also be shifted. In a word, it will get disturbed by that test charge. But the book derived the formula as if the positive point charge $+q$ remained fixed i.e. the test charge had no effect on the field of the point charge. But how can it be? If the test charge $q_0$ is a charge, then obviously it has also field which will exert force on $+q$. So how can the formula be true? Why did the book neglect this?
This phenomenon can be compared to earth's gravitational field. When a body moves upward, it exerts force on the earth downward. But as the earth has huge mass, its shifting can be neglected and hence it doesn't affect its gravitational field. 
But here it is a point positive charge; it has no huge mass. So why did the book neglect its shifting and how can the formula be true? Please help.  
 A: The presence of the test particle may influence the distribution of charge in the charged object. The electric field measured the above way then depends on the quantity of charge in the test particle and is different from what it is when the test particle is removed.
If the charge of the test particle is much lower than the charge of the charged object, this influence becomes weak and the difference negligible.
A: The definition of Electric Field is: 

The electric field at a point is defined as the limit of the ratio between the force on a test charge placed at that point to the test charge as the test charge approaches to zero. 

So in the definition, we have seen that the placed test charge is vary small and hence the effect of this charge on the field is also very small. This small effect of test charge on the field may be neglected. $$\vec{E} (\vec{r}) = \lim_{q \to 0} (\frac{\vec{F}}{q})$$ .
A: It is called the electrostatic field because all the movement of the charges and time dependence has been ignored, hence static.
On the other hand, when you place two charges $q_0$ and $+q$ they will both indeed create a field. However, what you measure is the force acting on $q_0$ and since a charge is not affected by it's own field then this wouldn't be counted in $\vec{E}$.
A: Due to the superposition principle,there is no movement of the test charge.And also there is no need of using the limit to define the electrostatic field.
