What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1) ?

Question

What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1) ?

Answer ( Expert Verified )

Strength of the electric field:
E = k Q / d²
k = 8.99 · 10 ^9
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 2² = √ (5² + 10²) =√ 125 = 11.18 cm = 0.1118 m
E 2 = 8.99 · 10 ^9 · 3 · 10^(-9)C / ( 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.

Leave a Comment