Respuesta :
Answer:
Explanation:
Given
Charge [tex]q_1=13\ nC is placed at x=0[/tex]
Charge [tex]q_2=8\ nC is placed at x=5\ m[/tex]
Electric field because of both the charges will be away from them
Electric field because of charge [tex]q_1[/tex] at distance r from it
[tex]E_1=\frac{kq_1}{r^2}[/tex]
[tex]E_1=\frac{9\times 10^9\times 13\times 10^{-9}}{r^2}[/tex]
Electric Field due to charge [tex]q_2[/tex] at distance of 5-r from it
[tex]E_2=\frac{kq_2}{(5-r)^2}[/tex]
[tex]E_2=\frac{9\times 10^9\times 8\times 10^{-9}}{(5-r)^2}[/tex]
at this Point Net Electric field is zero i.e.
[tex]E_1=E_2[/tex]
[tex]\frac{9\times 10^9\times 13\times 10^{-9}}{r^2}=\frac{9\times 10^9\times 8\times 10^{-9}}{(5-r)^2}[/tex]
[tex]\frac{5-r}{r}=\sqrt{\frac{8}{13}}[/tex]
[tex]5-r=0.784 r[/tex]
[tex]5=1.784 r[/tex]
[tex]r=\frac{5}{1.784}[/tex]
[tex]r=2.80\ m[/tex]
Thus at [tex]x=2.8\ m[/tex] net electric field is zero
