Answer:
E = -3.6859 x 10∧6 N/C
Explanation:
Let q1 = 3.5 μF = 3.5 x 10∧-6 F and q2 =-7.6 μF = -7.6 x 10∧-6 F
are the charges placed at two corner of equilateral triangle. Electric Field Magnitude "E" on the third corner will be equal to the sum of Electric Filed Magnitude generated by q1 and q2.
E = E1 + E2
E = Ke × q1 / ([tex]d^{2}[/tex]) + Ke × q2 / ([tex]d^{2}[/tex])
E = Ke/[tex]d^{2}[/tex] (q1 + q2) (taking common)
(Ke 8.99 × 10∧9 N[tex]m^{2}[/tex]/[tex]C^{2}[/tex] and distance d= 0.1 m)
E = 8.99 × 10∧9 N[tex]m^{2}[/tex]/[tex]C^{2}[/tex] /[tex](0.1 m)^{2}[/tex] (3.5 x 10∧-6 F - 7.6 x 10∧-6 F)
E = -3685.9 x 10∧3 N/C
E = -3.6859 x 10∧6 N/C
