a. Cường độ điện trường:
$U_{BC}=E.d_{BC} \Rightarrow E=\dfrac{U_{BC}}{d_{BC}}=\dfrac{600}{0,2}=3000V/m$
• $U_{AC}$
$d_{AC}=AA=0$
$U_{AC}=E.d_{AC}=0$
• $U_{BA}$
$d_{BA}=BA=BC.cos60=0,2.cos60=0,1m$
$U_{BA}=E.d_{BA}=3000.0,1=300V$
b.
$d_{AB}=-0,1m$
$A_{AB}=qEd_{AB}=10^{-9}.3000.(-0,1)=-3.10^{-7}J$
$A_{BC}=qEd_{BC}=10^{-9}.3000.0,2=6.10^{-7}J$
$A_{AC}=qEd_{AC}=0$
c.
$E_{C}=k.\dfrac{|q|}{CA^2}=9.10^9.\dfrac{|9.10^{-10}|}{(0,2.sin60°)^2}=270V/m$
$\vec{E_A}=\vec{E_{C}}+\vec{E}$
$\vec{E_{C}}\perp\vec{E}$
$\Rightarrow E_A=\sqrt{E_{C}^2+E^2}=\sqrt{270^2+3000^2}=3012V/m$