A charged oil drop is suspended in uniform field of 3 × 104 V/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge = 9.9 × 10-15 kg and g = 10 m/s2)

Question

A charged oil drop is suspended in uniform field of 3 × 104 V/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge = 9.9 × 10-15 kg and g = 10 m/s2) (A) 3.3 × 10-18 C (B) 3.2 × 10-18 C (C) 1.6 × 10-18 C (D) 4.8 × 10-18 C

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/d59e0681f0ad28374952a272254e5978-.jpeg" /> Given: mass o f charged drop m=9.9× 10-15 kg Electric field E=3× 104 V/m In steady state. Electric force on a drop = weight of a drop ∴ qE=mg or q=(mg/E) =(9.9×10-15×10/3× 104) =3.3× 10-18 C

Q. A charged oil drop is suspended in uniform field of $3 \times 1 0^{4} V / m$ so that it neither falls nor rises. The charge on the drop will be
(Take the mass of the charge $= 9.9 \times 1 0^{- 15} k g$ and $g = 10 m / s^{2})$

$3.3 \times 1 0^{- 18} C$

$3.2 \times 1 0^{- 18} C$

$1.6 \times 1 0^{- 18} C$

$4.8 \times 1 0^{- 18} C$

Given : mass o f charged drop
$m = 9.9 \times 1 0^{- 15} k g$
Electric field $E = 3 \times 1 0^{4} V / m$
In steady state.
Electric force on a drop = weight of a drop
$∴ qE = m g$
or $q = \frac{m g}{E}$
$= \frac{9.9 \times 1 0 ^{- 15} \times 10}{3 \times 1 0 ^{4}}$
$= 3.3 \times 1 0^{- 18} C$

Q. A charged oil drop is suspended in uniform field of 3×104V/m so that it neither falls nor rises. The charge on the drop will be (Take the mass of the charge =9.9×10−15kg and g=10m/s2)

3.3×10−18C

3.2×10−18C

1.6×10−18C

4.8×10−18C