Respuesta :
The strength of the electric field is 2157 N/C, the angle at the net electric field is 9.33° above the horizontal.
Strength of the electric field:
= E = k Q / d²
Given;
k = 8.99 · 10⁹
d 1 = 5 cm = 0.05 m
Q 1 = 3 n C = 3 · 10⁻⁹ C
E 1 = 8.99 · 10^9 · 3 · 10⁻⁹ C / ( 0.05 m )² = 10,788 N / C
E 1 = 10,788 i + 0 j
(d 2)² = √ (5² + 10²) =√ 125 = 11.18 cm = 0.1118 m
E 2 = 8.99 · 10⁹ · 3 · 10⁻⁹C / ( 0.1118 m )² = 2,157 N/C
Angle:
cot⁻¹ ( 5/10 ) = cot⁻¹ 0.5 = 63.43°
= E 2 = cos 63.43° · 2,154 i + sin 63.43° · 2,154 j
= E 2 = 963.46 i + 1,927.02 j
E = E 1 + E 2 = 11,731.46 i + 1,927.02 j
= √ (11,731.46² + 1,927.02² )
= 11,888 N/C
α = tan⁻¹ ( 1,927.02/11,731.46) = tan⁻¹ 0.1643 = 9.33° above the horizontal.
-- The given question is incomplete, I answer the question in general according to my knowledge and the complete question is
"What are the strength and direction of the electric field at the position indicated by the dot in (figure 1)? Where, the two positive charges of 3 nC are at a distance of 10 cm and the distance between the first proton and the position indicated is 5 cm." --
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