Answer:
-9Q
Explanation:
Electric field at origin is:
[tex]E=\frac{2keQ}{a^2}[/tex]
Electric field due to first charge at origin would be:
[tex]E_1=\frac{keQ}{a^2}[/tex]
Electric field due to second charge would be:
[tex]E_2=E-E_1\\E_2=\frac{2keQ}{a^2}-\frac{keQ}{a^2} = \frac{keQ}{a^2}[/tex]
If the second charge is Q', then [tex]E_2[/tex] should be:
[tex]E_2=\frac{keQ'}{(3a)^2}=\frac{keQ'}{9a^2}[/tex]
compare the above two values to find the possible values of Q':
[tex]\frac{|Q'|}{9}=Q\\ |Q'|=9Q[/tex]
The net electric field at origin is greater than the one due to first charge. It means the second charge adds on to the electric field at the origin. Thus, it should be a negative charge.
Thus, Q' = -9Q
One value is possible as the location of the second charge is given to be on the positive x-axis.
