Difference between incident ray and refracted ray
1 answer:
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Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
Answer:
θ = cos^(-1) (-A/B)
Explanation:
The image of the reauktant forces A & B are missing, so i have attached it.
Now, from the attached image, we will see that;
Angle between A and B is θ
Also;
A = Bcos(180° − θ)
Now, in trigonometry, we know that;
cos(180° − θ) = -cosθ
Thus;
A = -Bcosθ
cosθ = -A/B
Thus;
θ = cos^(-1) (-A/B)
