Question

Marking scheme: +4 for correct answer, O if not attempted and in all other case
20. Atiny spherical oil drop carrying a net charge q is balanced in still air with a vertical
817
uniform electric field of strength
x10 V/m. When the field is switched off, the drop
is observed to fall with terminal velocity 2.10-mi's. Given g =98/s. viscosity of
the air 1.8x10-Ns/m and the density of oil 900kg/m neglect air resistance and
conduction between oil drop and air and the magnitude of q. is:
10-19​

Answer 1

The magnitude of charge on the drop is q = 8.0×10^−19 C

Explanation:

qE=mg    .......(1)  

6πηrv = mg

43πr^3ρg = mg   .......(2)

r =( 3mg / 4πρg)^1/3      ........(3)

Substituting the value of "r" in Equation (ii) we get,

6πηv(3mg / 4πρg)^1/3 = mg

or (6πηv)^3 (3mg / 4πρg)=(mg)^3

Substituting mg = qE we get,

(qE)^2 = (3 / 4πρg)(6πηv)^3

or qE=(3 / 4πρg)^1/2(6πηv)^3/2

q = 1 / E(3 / 4πρg)^1/2(6πηv)^3/2

Substituting the values we get,

q = 7 / 81π×105 √ 3/4π×900×9.8×216π^3 ×√(1.8×10^−5×2×10^−3)^3

q = 8.0×10^−19 C

Thus the magnitude of charge on the drop is q = 8.0×10^−19 C

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