✔ 最佳答案
1. Find the electric field at a point midway between two chatges of +30*10-9 C and +60*10-9 C seperated by a distance og 30cm
E = Q1 / (4πεoR2) – Q2 / (4πεoR2)
= (60x10-9 – 30x10-9) / (4π8.85x10-12(0.15)2)
= 11989NC-1 (direction to the 30x10-9 C)
2. A +5.7 uC point charge is on the x-axis at x = -3m, and a +2uC point charge is on the x-axis at x = +1m. Determine the net electric field (magnitude and direction) on the y-axis at y = +2m.
from the graph
圖片參考：
http://hk.geocities.com/namsm4e/P11.jpg
E1 = Q1 / (4πεoR12) = 5.7x10-6 / (4π(8.85x10-12)( √13)2)
= 3942.6NC-1
E2 = Q2 / (4πεoR22) = 2.0x10-6 / (4π(8.85x10-12)( √5)2)
= 3596.7NC-1
sinα = 2/√13 sinβ = 2/√5
cosα = 3/√13 sinβ = 1/√5
ΣEx = 3942.6cosα + 3596.7cosβ
= 3942.6x3/√13 - 3596.7x1/√5
= 1671.9
ΣEy = 3942.6sinα + 3596.7sinβ
= 3942.6x2/√13 + 3596.7x2/√5
= 5403.9
E = √(ΣEx2 + ΣEy2)
= √(1671.92 + 5403.92)
= 5656.7NC-1
θ = tan-1(5403.9/1671.9)
= 72.8o
