r₁ = distance of point A from charge q₁ = 0.13 m
r₂ = distance of point A from charge q₂ = 0.24 m
r₃ = distance of point A from charge q₃ = 0.13 m
Electric field by charge q₁ at A is given as
E₁ = k q₁ /r₁² = (9 x 10⁹) (2.30 x 10⁻¹²)/(0.13)² = 1.225 N/C towards right
Electric field by charge q₂ at A is given as
E₂ = k q₂ /r₂² = (9 x 10⁹) (4.50 x 10⁻¹²)/(0.24)² = 0.703 N/C towards left
Since the electric field in left direction is smaller, hence the electric field by the third charge must be in left direction
Electric field at A will be zero when
E₁ = E₂ + E₃
1.225 = 0.703 + E₃
E₃ = 0.522 N/C
Electric field by charge "q₃" is given as
E₃ = k q₃ /r₃²
0.522 = (9 x 10⁹) q₃/(0.13)²
q₃ = 0.980 x 10⁻¹² C = 0.980 pC
Answer:
low, low
Explanation:
Longer wavelengths will have lower frequencies, and shorter wavelengths will have higher frequencies.
Large amplitude waves contain more energy. The other is frequency, which is the number of waves that pass by each second. If more waves( or more wiggly lines) pass by, more energy is transferred each second
Hmm, I will come back to this one just to help. :)