A tiny spherical oil drop carrying a net charge q is balanced in still air with a vertical uniformelectric field of strength 𝐸𝑉 𝑚−1. When the field is switched off, the drop is observed to fall witha terminal velocity of 𝑣 𝑚/𝑠. Given 𝑔 = 9.8 𝑚/𝑠2, viscosity of the air = 𝜂 𝑁 𝑠𝑚−2 and the densityof the oil = 𝜌 𝑘𝑔/𝑚3, find the magnitude of 𝑞

To find the magnitude of the charge q, we need to consider two situations: when the electric field is on and when it is off.

1. When the electric field is on, the oil drop is balanced. This means the electric force (F_e) is equal to the gravitational force (F_g). We can express this as:

   F_e = F_g qE = mg q = mg/E ----(1)

2. When the electric field is off, the oil drop falls with a terminal velocity. At terminal velocity, the gravitational force is balanced by the viscous drag (F_d). We can express this as:

   F_g = F_d mg = 6πηrv r = mg/(6πηv) ----(2)

   Where r is the radius of the oil drop, η is the viscosity of air, and v is the terminal velocity.

3. We also know that the mass m of the oil drop can be expressed in terms of its density (ρ) and volume (which for a sphere is 4/3πr³). So:

   m = ρ(4/3πr³) r = (m/(ρ(4/3π)))^(1/3) ----(3)

4. Equating the expressions for r from (2) and (3) gives us:

   mg/(6πηv) = (m/(ρ(4/3π)))^(1/3) Solving for m gives: m = (9vη/(2gρ))^(1/2)

5. Substituting m from (4) into (1) gives:

   q = ((9vη/(2gρ))^(1/2) * g) / E Simplifying gives: q = (9vηg/(2E²ρ))^(1/2)

So the magnitude of the charge q on the oil drop is given by the square root of the product of 9 times the terminal velocity, the viscosity of air, and the acceleration due to gravity, divided by twice the product of the electric field strength squared and the density of the oil.