Respuesta :
In BPC
tan\theta =a/b = 3/4
\theta = tan^-1(0.75)
\theta = 36.87 deg
BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m
Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C
Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C
Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C
Net electric field along X-direction is given as
Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C
Net electric field along X-direction is given as
Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C
Net electric field is given as
E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C
tan\theta =a/b = 3/4
\theta = tan^-1(0.75)
\theta = 36.87 deg
BP = sqrt(a^2 + b^2) = sqrt((3)^2 + (4)^2) = 5 m
Eb = k Q/BP^2 = (9 x 10^9) (16 x 10^-9)/5^2 = 5.76 N/C
Ea = k Q/AP^2 = (9 x 10^9) (16 x 10^-9)/4^2 = 9 N/C
Ec = k Q/CP^2 = (9 x 10^9) (16 x 10^-9)/3^2 = 16 N/C
Net electric field along X-direction is given as
Ex = Ea + Eb Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C
Net electric field along X-direction is given as
Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C
Net electric field is given as
E = sqrt(Ex^2 + Ey^2) = sqrt((13.6)^2 + (19.5)^2) = 23.8 N/C
The magnitude of the electric field will be equal to E=23.8 N/C
What is electric field?
Electric field, an electric property associated with each point in space when charge is present in any form.
The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field
In BPC
[tex]tan\theta =\dfrac{a}{b} =\dfrac{ 3}{4}[/tex]
[tex]\theta = tan^-1(0.75)[/tex]
[tex]\theta = 36.87 deg[/tex]
[tex]BP = \sqrt{(a^2 + b^2)} = \sqrt{(3)^2 + (4)^2)} = 5 m[/tex]
[tex]E_b = \dfrac{k Q}{BP^2} = \dfrac{(9 \times 10^9) (16 \times 10^-9)}{5^2} = 5.76 N/C[/tex]
[tex]E_a = \dfrac{k Q}{AP^2} = \dfrac{(9 \times 10^9) (16 \times 10^-9)}{4^2} = 9 N/C[/tex]
[tex]E_c = \dfrac{k Q}{CP^2} = \dfrac{(9 \times 10^9) (16 \times 10^-9)}{3^2} = 16 N/C[/tex]
Net electric field along X-direction is given as
[tex]Ex = Ea + E_b Cos36.87 = (9) + (5.76) Cos36.87 = 13.6 N/C[/tex]
Net electric field along y-direction is given as
[tex]Ey = Ec + Eb Sin36.87 = (16) + (5.76) Sin36.87 = 19.5 N/C[/tex]
Net electric field is given as
[tex]E = \sqrt{(Ex^2 + Ey^2)} = \sqrt{(13.6)^2 + (19.5)^2)} = 23.8 N/C[/tex]
Hence the magnitude of the electric field will be equal to E=23.8 N/C
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