Question

Answer this Question of iit jee .
Chapter → Electrostatics​​

Answer 1

EXPLANATION.

A charged oil drop is suspended in a uniform field.

⇒ 3 x 10⁴ V/m.

As we know that,

Mass of charge = 9.9 x 10⁻¹⁵ kg.

⇒ g = 10 m/s².

⇒ qE = mg.

⇒ q = mg/E.

⇒ q = 9.9 x 10⁻¹⁵ x 10/3 x 10⁴.

⇒ q = 9.9 x 10⁻¹⁴/3 x 10⁴.

⇒ q = 9.9/3 x 10⁻¹⁴ x 10⁻⁴.

⇒ q = 3.3 x 10⁻¹⁸C.

Option [A] is correct answer.

Answer 2

Answer:

A charged oil drop is suspended in a uniform feild so that it neither falls nor rises

So,

Force on oil drop = Gravitational force applied on oil drop

given,

E = 3×10⁴V/m

$m = 9.9 \times {10}^{ - 15} kg$

g = 10 m/s²

$→qE = mg$

$→q \times 3 \times {10}^{4} = 9.9 \times {10}^{ - 15} \times 10$

$→q = \frac{9.9 \times {10}^{ - 15} \times 10}{3 \times {10}^{4} }$

$→q = 3.3 \times \times {10}^{ - 18} C$