A machine which has an energy loss of 10% will have efficiency of
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Answer:
Magnitude of the resulting force on the 7 nC charge at the origin:
Fn₁= 23.95*10⁻⁹ N
Explanation:
Look at the attached graphic:
Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.
q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:
F = (k*q₁*q)/(d)²
m : distance from q₁ to q₂
(d₁₂)² = 18 m²
m : distance from q₁ to q₃
(d₁₃)² = 10 m²
m : distance from q₁ to q₄
(d₁₄)² = 13 m²
K= 8.98755 × 10⁹ N *m²/C²
q₁= 7*10⁻⁹C
k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9
F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C
F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C
F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C
x-y components of the net force on q₁ (Fn₁):
α= tan⁻¹(3/3)= 45° , β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°
Fn₁x = F₂₁x+ F₃₁x+F₄₁x
F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N
F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N
F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N
Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N
Fn₁x = -23.87 *10⁻⁹ N
Fn₁y = F₂₁y+ F₃₁y+F₄₁y
F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N
F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N
F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N
Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N
Fn₁y = 1.97*10⁻⁹ N
Magnitude of the resulting force on the 7 nC charge at the origin (q₁):
Fn₁= 23.95*10⁻⁹ N
