A charged oil drop is suspended in a 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 a 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

Since ball is hanging in equilibrium, force by gravity is balanced by electric force. qE = mg ⇒ q = (m× g/E) ⇒ (9.9× 10-15× 10/3× 104) ∴ q = 3.3× 10-18 C

Q. A charged oil drop is suspended in a 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 $g = 10 m / s^{2}$ )

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

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

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

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

Since ball is hanging in equilibrium, force by gravity is balanced by electric force.
$qE = m g$
$\Rightarrow q = \frac{m \times g}{E}$
$\Rightarrow \frac{9.9 \times 1 0 ^{- 15} \times 10}{3 \times 1 0 ^{4}}$
$∴ q = 3.3 \times 1 0^{- 18} C$

Q. A charged oil drop is suspended in a 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.