Answer:
[tex]E_{net} = 2.25 KQ\L2[/tex]
Explanation:
from figure
[tex]d = \sqrt { l^2 -(\frac{l}{2})^2}[/tex]
[tex]= \sqrt {\frac{3l^2}{4}[/tex]
[tex]= l\sqrt{ \frac{3}{4}}[/tex]
net field at point P
[tex]E_{net} = E_1 -E_2 +E_3[/tex]
[tex]= K\frac{q_1}{(\frac{l}{2})^2} -K\frac{q_2}{(\frac{l}{2})^2} +K\frac{q_3}{d^2}[tex]
[tex]q_1 =q_2 =q_3 =27Q[/tex]
[tex]d = l\sqrt{ \frac{3}{4}}[/tex]
[tex]E_{net} = K\frac{27Q}{(\frac{l}{2})^2} -K\frac{27Q}{(\frac{l}{2})^2} +K\frac{27Q}{(l\sqrt{ \frac{3}{4}})^2}[/tex]
[tex]E = \frac{27 KQ}{\frac{3}{4}l^2}[/tex] {l =4l}
[tex]E_{net} = \frac{4 *36 KQ}{3{4l}^2}[/tex]
[tex]E_{net} = 2.25 KQ\L2[/tex]
