Question

A +4.0

Answer 1

in triangle ABC ,

AB = 4 cm

AC = 3 cm

Using pythagorean theorem :

BC² = AB² + AC²

BC² = 3² + 4²

BC = 5 cm

electric field by a charge is given as ::

electric field by the negative charge at C is given as ::

E₁ = kq₁/r²₁

r₁ = 3 cm = 0.03 m

E₁ = (9 x 10⁹)(3 x 10^-9)/(0.03)²

E₁ = 30000 N/C   in negative Y-direction

electric field by the positive charge at C is given as ::

E₂ = kq₂/r²₂

r₂ = 5 cm = 0.05 m

E₂ = (9 x 10⁹)(4 x 10^-9)/(0.05)²

E₂ = 14400 N/C

the components of E₂ in X- and Y-direction is given as ::

E_2x = E₂ Cos= 14400 (AB/BC) = 14400 x (4/5) = 11520 N/C

E_2y = E₂ Sin= 14400 (AC/BC) = 14400 x (3/5) = 8640 N/C    in positive y-direction

net force in Y-direction

E_y = E₁ - E_2y= 30000 - 8640 = 21360 N/C

net force in X-direction

E_x = E_2x= 11520 N/C

net electric field = E_net= sqrt (E_x² + E_y²) = sqrt (11520² + 21360²) = 24268.5 N/C

b)

tan' = E_y/E_x = 21360/11520

' = 61.7