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Question: Two point charges \[q\] and \[ - 2q\] are placed some distance apart. If the electric field at the l...

Two point charges qq and 2q - 2q are placed some distance apart. If the electric field at the location of qq due to 2q - 2q is EE, than at the location of 2q - 2q due to qq is ( No external Field is present )
A. E2 - \dfrac{E}{2}
B. 2E - 2E
C. E2\dfrac{E}{2}
D. 4E - 4E

Explanation

Solution

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 qq is given by,
E=14πε0qr2r^\vec E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}\hat r
where, rr is the distance from the charge and ε0{\varepsilon _0} is the electric permittivity of the medium is the unit vector along rr from the charge.

Complete step by step answer:
We have given here two charges qq and 2q - 2q placed at some distance. The electric field due to 2q - 2q at the location of qq is EE. 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 qq is given by,
E=14πε0qr2r^\vec E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}\hat r
Magnitude of the field will be,
E=14πε0qr2E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}.
So, the electric field due to charge 2q - 2q at the location of qqis EE which is equal to, E=14πε02qr2E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{ - 2q}}{{{r^2}}}
Now, the electric field due to the charge qq at the position of 2q - 2q is,
E1=14πε0qr2{E_1} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}
So, we can write, E=14πε02qr2E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{ - 2q}}{{{r^2}}}
12E=14πε0qr2- \dfrac{1}{2}E = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{q}{{{r^2}}}
E1=E2\therefore {E_1} = - \dfrac{E}{2}
So, the electric field due to the charge qq at the position of 2q - 2q is E2 - \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.