In This post you will get answer of Q.
Two charges $q $ and $-2q$ are separated by a distance $d$. If the electric intensity at the site of $q$ is $E$ then electric field at the site of $-2q$ is ·
Solution:
Given situation is shown in the figure,
Also, net electric field at the site of charge q is E.
$E = frac{-2q }{4 pi varepsilon_0 d^2} , , , , , , , , , , , , , , , , , $ …(i)
Net electric field at the site of $-2q$ is $E’$ (say).
$therefore : E = frac{q}{4 pi varepsilon_0 d^2} , , , , , , , , , , , , , , , , , $ …(ii)
From eq. (i) and (ii),
$E’ = – frac{E}{2}$

Solution:
Given situation is shown in the figure,
Also, net electric field at the site of charge q is E.
$E = frac{-2q }{4 pi varepsilon_0 d^2} , , , , , , , , , , , , , , , , , $ …(i)
Net electric field at the site of $-2q$ is $E’$ (say).
$therefore : E = frac{q}{4 pi varepsilon_0 d^2} , , , , , , , , , , , , , , , , , $ …(ii)
From eq. (i) and (ii),
$E’ = – frac{E}{2}$