Question

In an oil drop experiment, 'n' numbers of electrons are stripped from an oil drop to make it positively charged. A vertical electric field of magnitude 4.9x10¹⁴ N/C is applied to balance the force due to gravity on the oil drop. If the mass of oil drop is 80 μg,the value of 'n' will be:(Take g =9.8 m/s² and charge of an electron = 1.6 x 10^-19 C)

• A. 1
• B. 10
• C. 100
• D. 1000
• E. 10000

Answer 1

The Correct Option is B

 The correct answer is (B):10

Answer 2

Let's Convert the mass of oil drop to kilograms.
Given: Mass of oil drop = 80 μg = 80 × 10^-6 kg

Calculate force due to gravity.
Given: Gravity (g) = 9.8 m/s²
Formula: Force gravity = mass of oil drop \(\times\) gravity
Force gravity = \(80 \times 10^{-6} \, \text{kg} \times 9.8 \, \text{m/s}^2\)
\(=\gt 7.84 \times 10^{-4} \, \text{N}\)

Balance forces using an electric field.
Given: Electric field strength \(E = 4.9 \times 10^{14}\ N/C\)

Solve for total charge (q).
Formula: \(\text{q} = \frac{\text{force\_gravity}}{\text{E}}\)

\(\text{q} = \frac{7.84 \times 10^{-4} \, \text{N}}{4.9 \times 10^{14} \, \text{N/C}}\)

\(\text{q} = 1.6 \times 10^{-18} \, \text{C}\)

the number of missing electrons (n).
Given: Charge of a single electron = \(1.6 \times 10^{-19}\ C\)
Formula: \(\text{n} = \frac{\text{q}}{\text{charge\_of\_electron}}\)

\(\text{n} = \frac{|1.6 \times 10^{-18} \, \text{C}|}{1.6 \times 10^{-19} \, \text{C}}\)
\(\text{n} = 10\)

So, the correct option is (B): 10