Question

An oil drop has a charge $-9.6 \times 10^{-19}C$ and mass $1.6 \times 10^{-15}$ gm. When allowed to fall, due to air resistance force it attains a constant velocity. Then if a uniform electric field is to be applied vertically to make the oil drop ascend up with the same constant speed, which of the following are correct. ($g = 10ms^{-2}$) (Assume that the magnitude of resistance force is same in both the cases)

• A. The electric field is directed upward
• B. The electric field is directed downward
• C. The intensity of electric field is $\frac{1}{3} \times 10^2 NC^{-1}$
• D. The intensity of electric field is $\frac{1}{6} \times 10^5 NC^{-1}$

Answer 1

Answer: (B), (C)

B, C

Answer 2

Let the mass of the oil drop be $m$ and its charge be $q$. The acceleration due to gravity is $g$.

Given $m = 1.6 \times 10^{-15}$ gm $= 1.6 \times 10^{-15} \times 10^{-3}$ kg $= 1.6 \times 10^{-18}$ kg. Given $q = -9.6 \times 10^{-19}$ C. Given $g = 10$ m/s$^2$.

Case 1: Oil drop falling with constant velocity.

When the oil drop falls with a constant velocity, the net force on it is zero. The forces acting are the gravitational force downwards ($F_g = mg$) and the air resistance force upwards ($F_r$). Since the velocity is constant, the net force is zero: $F_r - F_g = 0$, so $F_r = F_g = mg$. $F_r = (1.6 \times 10^{-18} \text{ kg}) \times (10 \text{ m/s}^2) = 1.6 \times 10^{-17}$ N.

Case 2: Oil drop moving upward with the same constant velocity.

When the oil drop moves upward with the same constant velocity, the net force on it is also zero. The forces acting are the gravitational force downwards ($F_g = mg$), the air resistance force downwards ($F_r$, since the motion is upward), and the electric force ($F_e$).

For the oil drop to move upward, the electric force must be directed upward. The charge of the oil drop is negative ($q < 0$). The electric force is given by $F_e = qE$. Since $q$ is negative, for $F_e$ to be upward, the electric field $E$ must be directed downward.

Let's take the upward direction as positive. The net force is $F_e - F_g - F_r = 0$. So, $F_e = F_g + F_r$. From Case 1, we know $F_r = mg$. Thus, $F_e = mg + mg = 2mg$. The magnitude of the electric force is $|F_e| = |qE| = |q||E|$. So, $|q||E| = 2mg$. The magnitude of the electric field is $|E| = \frac{2mg}{|q|}$.

Substitute the given values: $|E| = \frac{2 \times (1.6 \times 10^{-18} \text{ kg}) \times (10 \text{ m/s}^2)}{|-9.6 \times 10^{-19} \text{ C}|}$ $|E| = \frac{2 \times 1.6 \times 10^{-17}}{9.6 \times 10^{-19}} \frac{\text{N}}{\text{C}}$ $|E| = \frac{3.2 \times 10^{-17}}{9.6 \times 10^{-19}} \frac{\text{N}}{\text{C}}$ $|E| = \frac{3.2}{9.6} \times 10^{-17 - (-19)} \frac{\text{N}}{\text{C}}$ $|E| = \frac{1}{3} \times 10^2 \frac{\text{N}}{\text{C}}$.

Therefore, options B and C are correct.