The general formula is;
Pressure = Force/Area
Where,
Pressure = Required pressure + Atmospheric pressure = (1.2*10^5) + (101325) = 221325 Pa = 221325 N/m^2
Area = πD²/4 = π*0.035²/4 = 9.621*10^-4 m²
Thus,
Force, F = Pressure*Area = 221325*9.621*10^-4 = 212.94 N
Response:
Magnitude of the electrostatic force acting on the +32 µC charge, 
Clarification:
Let q₁ = +32 µC, located at x₁ = 0
q₂ = +20 µC, positioned at x₂ = 40 cm = 0.4 m
q₃ = -60 µC, placed at x₃ = 60 cm = 0.6 m
Define the force magnitude on the +32 µC charge from the +20 µC charge as F₁ (the force on q₁ due to q₂).


Define the force magnitude on the +32 µC charge from the -60 µC charge as F₂ (the force on q₁ due to q₃).


The resultant electrostatic force on the 32 µC charge is 

The question Ellen is likely exploring is "In what way does distance influence the gravitational force acting on objects?"
Explanation:
For a string that is secured at both ends, the oscillation amplitude must be zero at both extremities of the string. Refer to the attached images for clarity. It is evident that our fundamental harmonic will have the following wavelength: All subsequent harmonics are simply multiples of the fundamental. The three longest wavelengths are as follows: