Answer:
net electrostatic force = 46.76 N
and direction is vertical downward
Explanation:
given data
charge = +5μC
corners identical charges = -6 μC
length of side = 0.10 m
to find out
What is the net electrostatic force
solution
here apex is the vertex where the two sides of equal length meet
so upper point A have charge +5μC and lower both point B and C have charge -6 μC
so
force between +5μC and -6 μC is express as
force = [tex]k \frac{q1q2}{r^2}[/tex] ..............1
put here electrostatic constant k = 9 × [tex]10^{9}[/tex] Nm²/C² and q1 q2 is charge given and r is distance 0.10 m
so
force = [tex]9*10^{9} \frac{5*6*10^(-12)}{0.10^2}[/tex]
force = 27 N
so net force is vector addition of both force
force = [tex]\sqrt{x^{2}+x^{2}+2x^{2}cos60}[/tex]
here x is force 27 N
force = [tex]\sqrt{27^{2}+27^{2}+2(27)^{2}cos60}[/tex]
net electrostatic force = 46.76 N
and direction is vertical downward
