Answer:
Explanation:
Given
[tex]q_1=11.5\ nC[/tex] charge is placed at [tex]x=0\ cm[/tex]
another charge of [tex]q_2=-1.2\ nC[/tex] is at [tex]x=3\ cm[/tex]
We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.
so net electric field is zero somewhere beyond negatively charged particle
Electric Field due to [tex]q_2[/tex] at some distance r from it
[tex]E_2=\frac{kq_2}{r^2}[/tex]
Now Electric Field due to [tex]q_1[/tex] is
[tex]E_1=\frac{kq_1}{(3+r)^2}[/tex]
Now [tex]E_1+E_2=0[/tex]
[tex]\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0[/tex]
[tex]\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}[/tex]
[tex]\frac{3+r}{r}=3.095[/tex]
thus [tex]r=1.43\ cm[/tex]
Thus Electric field is zero at some distance r=1.43 cm right of [tex]q_2[/tex]
