Does any object placed in an electric field change the electric field? Lets say I have a point charge of magnitude $+q$, All around it I would have a symmetric radial electric field. Now if I place a neutral object lets say a sphere (doesn't matter insulating or conducting) in this field some distane away from the point charge. A negative charge will be induced on the object near the point charge and a positive charge on the opposite side.
No matter how small this induced charge is, due to the radial distance of the two (positive and negative) there must be an increase/decrease in net electric field on either side of the object and mostly everywhere else too !
I hope that what I am thinking is wrong, because we have not been taught that anything placed in electric field would affect the field itself regardless of it's nature. But I can't figure out what am I thinking wrong, how to solve this dilemma ?
 A: If the material placed in the field of the positive charge is a conductor, the field will be distorted and the method to see the field  is the image charges  method. It will depend on the boundary conditions. 
For a grounded conducting sphere


Field lines outside a grounded sphere for a charge placed outside a sphere.

For a non grounded conductor:


This illustration shows a spherical conductor in static equilibrium with an originally uniform electric field. Free charges move within the conductor, polarizing it, until the electric field lines are perpendicular to the surface. The field lines end on excess negative charge on one section of the surface and begin again on excess positive charge on the opposite side. No electric field exists inside the conductor, since free charges in the conductor would continue moving in response to any field until it was neutralized.

If the field is created by a point charge the geometry will change but the physics is the same.
If you have a positive point charge and bring into its field a dielectric, then the field lines will change  again depending on constants as :


Figure 6.6.6 Electric field intensity in and around dielectric rod of Fig. 6.6.5 for (a) e_b > a and (b) e_b< e_a.

One can again imagine the geometric changes for a field from a sphere.
In summary, the field distorts with the presence of matter, differently for a conductor or dielectric .
A: I'm not quite sure I understand why you have a problem with this - every static charge is a source or a drain of the electric field, depending on its sign. So obviously the field of a single charge at the origin will be different from the field of three charges or any other configuration.
The electric potential of such an ensemble of pointlike charges $q_i$ at a specific point, measured by an uncharged observer, will be $$\Phi(\vec r) = \frac{1}{4 \pi \epsilon_0} \cdot \sum_i \frac{q_i}{\left| \vec r - \vec r_i \right|}$$
At this point, instead of evaluating the sum you can do a multipole expansion $$\Phi(\mathbf r) = \frac{1}{4 \pi \epsilon_0} \left( \frac{Q}{r} + \frac{\mathbf r \cdot \mathbf p}{r^3} + \frac{1}{2} \sum_{k,l} Q_{kl} \frac{r_k \cdot r_l}{r^5}+... \right),$$which is essentially a Taylor series of ${\left| \vec r - \vec r_i \right|}^{-1}$. From there, you will get the electric field by taking the gradient of the approximated potential.
A: The right answer is this: when a charged point is placed in the electric field that is generated by other charged objects other than the charged particle in question. Then we can use the formula $F=qE$ to calculate the force acted on the charged point. But not all charged particles with finite size can be approximated as a "charged point."
