Question

Two point charges \[q\] and \[ - 2q\] are placed some distance apart. If the electric field at the location of \[q\] due to \[ - 2q\] is \[E\], than at the location of \[ - 2q\] due to \[q\] is ( No external Field is present )
A. \[ - \dfrac{E}{2}\]
B. \[ - 2E\]
C. \[\dfrac{E}{2}\]
D. \[ - 4E\]

Answer 1

Hint: The electric field due to a point charge is proportional to the charge of value of the charge and inversely proportional to the square of the distance between the point and the charge. The direction of it depends on the type of the charge.

Formula used:
The electric field due to a charge \[q\] is given by,
\[\vec E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}\hat r\]
where, \[r\] is the distance from the charge and \[{\varepsilon _0}\] is the electric permittivity of the medium is the unit vector along \[r\] from the charge.

Complete step by step answer:
 We have given here two charges \[q\] and \[ - 2q\] placed at some distance. The electric field due to \[ - 2q\] at the location of \[q\] is \[E\]. Now, we know that the electric field is forced due to charge on a unit positive charge placed at some distance. The expression of electric field due to a charge \[q\] is given by,
\[\vec E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}\hat r\]
Magnitude of the field will be,
\[E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}\].
So, the electric field due to charge \[ - 2q\] at the location of \[q\]is \[E\] which is equal to, \[E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{ - 2q}}{{{r^2}}}\]
Now, the electric field due to the charge \[q\] at the position of \[ - 2q\] is,
\[{E_1} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}\]
So, we can write, \[E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{ - 2q}}{{{r^2}}}\]
\[ - \dfrac{1}{2}E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}\]
\[\therefore {E_1} = - \dfrac{E}{2}\]
So, the electric field due to the charge \[q\] at the position of \[ - 2q\] is \[ - \dfrac{E}{2}\]

Hence, option A is the correct answer.

Note: The electric field due to one charge at the position of the other charge is different since the electric field is the magnitude of the strength of the force field and is irrespective of the body on which it is acting. But the force due to one charge on the other charge is always the same as Newton’s third law states that if a body exerts some force on another body the other body will exert an equal and opposite amount of force.