Question

In the figure the three particles are fixed in place and have charges q₁ = q₂ = +3e and q₃ = +3e. Distance a = 4.86 µm. What is the magnitude of the net electric field at point P due to the particles?

Answer 1

here,

q1=q2=q3=3e

and

a=4.86 um

let

electric field due to q1 is E1

electric field due to q2 is E2

electric field due to q3 is E3

now,

electric field E_net=E1-E2+E3

but,E1=E2

electric field E_net=E3

E_net=k*3e/r^2

from the diagram r=a/sqrt(2)

E_net=k*3e/(a/sqrt(2))^2

=k*6e/a^2

=9*10^9*6*1.6*10^-19/(4.86*10^-6)^2

=365.79 N/C