A positively charged oil drop of mass is injected into a region of uniform electric field between two oppositely charged, horizontally oriented plates spaced 16 cm apart. If the mass of the drop is 8.0 10-15 kg and it remains stationary when the potential difference between the plates is 2.44 kV, what is the magnitude of the charge on the drop

the magnitude of the charge on the drop is 5.163 × 10⁻¹⁸  C

Explanation:

Given that;

mass of the drop m = 8.0×10⁻¹⁵ kg

distance d = 16 cm = 0.16 m

potential difference between the plates V = 2.44 kV = 2440 v

acceleration of gravity g = 9.81 m/s²

the magnitude of the charge on the drop = ?

weight is balanced by the electrostatic force

weight = mg = 8.0×10⁻¹⁵ kg × 9.81 m/s² = 7.848 × 10⁻¹⁴

we know that; V = Ed

E = V/d =  2440 / 0.16 = 15200 v/m

Electrostatic force = qE

so weight = qE

q = weight / E

q = 7.848 × 10⁻¹⁴ / 15200

q = 5.163 × 10⁻¹⁸  C

Therefore, the magnitude of the charge on the drop is 5.163 × 10⁻¹⁸  C

diagram is missing, but i will try to explain your doubt.

answer:

amplitude is the maximum displacement or distance moved on a point of a vibrating body or wave measured from its equilibrium position.it is equal to half the length of vibration path.