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
For the electric field is measured for points at distances r,Electric field is mathematically given as
E=19.9*10^{-5}c/m^3
<h3>What is the
Electric field?</h3>
Generally, the equation for the electric field uniform charged is mathematically given as

Therefore
E=lr/3E0=3Ee0/r
Therefore
E=3*6*10^4*8.85*10^{-12}
E=19.9*10^{-5}c/m^3
In conclusion, E=19.9*10^{-5}c/m^3
E=19.9*10^{-5}c/m^3
Read more about Electric field
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CQ
Complete question attached below
Answer:
This is net charge on the surface is Q = σ₀ x (y + 2by²)
Explanation:
The surface charge density is defined as the amount of charge Q per unit area A
σ = dq / dA
dq = σ dA
Since the surface is a rectangular region we use an xy coordinate system so the area difference
dA = dxdy
dq = σ dx dy
We replace, evaluate the integral
∫ dq = ∫ σ₀ (1 + yb) dxdy
realizamos laintegral de dx
Q -0 =σ₀ ∫ (1 + yb) (x-0) dy
Where we evaluate We must recognize that the charge Q must be zero by the time X = 0 and Y = 0. At the starting point Q = 0 for x = 0
We perform the other integral (dy)
Q = σ₀ x (y + 2y² b)
Evaluated between Y = 0 and Y = y
Q = σ₀ x (y + 2by²)
This is net charge on the surface