The point charges $Q$ and $-2Q$ are placed some distance apart. If the electric field intensity at $Q$ is $E$ then what is the electric field at $-2Q$.
A) $\dfrac{E}{2}$
B) $E$
C) $2E$
D) $1.5E$

Hint: Electric field is the physical field that surrounds each electric charge and exerts force on all other charges in the field, which is given by:
$E = \dfrac{F}{q}$ (F is the force exerted by charge on other charges and q is the charge).
Electric field intensity is the force that would be experienced by unit test charge placed at a location. Using the above relations we will find the electric field intensity at $-2Q$ charge.

Complete step by step solution:
Let us first define more about Electric field and Electric field intensity.
Electric field: A measure of the force exerted by one charged body on another. Imaginary lines force or electric field lines of force or electric field originate on positive charges and terminate on negative charges. Electric field lines are thought of as elastic lines which repel each other in a direction perpendicular to the line itself.
Electric Field Intensity: Electric field intensity at any location is the force that would be experienced by unit test charge placed at location. A uniform electric field is an ideal case in which electric field lines are parallel with one another.
Now, we will do the calculation part of the problem.
We are given that electric field intensity given by charge $Q$ is $E$.
We are supposed to find the electric field intensity due to charge $-2Q$.
As we have mentioned in the hint the electric field is given by the formula:
$E = \dfrac{F}{Q}$...............(1)
In equation 1 if we replace $q$ by $2Q$, then electric field intensity is given by:
$E = \dfrac{F}{{2Q}}$............(2)
On comparing equation 1 and 2 we can get
Electric field intensity by charge $-2Q$ is $\dfrac{E}{2}$.

Therefore, option (A) is correct.

Note: There are many applications of electric field which are: usage in tissue engineering for measuring cell and tissue properties, electro spinning(use high voltage which causes huge amounts of electric field in the capillary tubes),used in Van de Graff generators, escalators etc.