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. 100
• C. 10
• D. 1000
• E. 10000

Answer 1

The Correct Option is B

1. Forces involved:

• Gravity (F_g): Pulling downwards. F_g = m \(\times\) g
• Electric Force (F_e): Pushing upwards. F_e = q \(\times\) E

Since the forces are balanced, F_g = F_e.

2. Variables:

• m (mass): 80 µg = 80 × 10^-9 kg
• g (acceleration due to gravity): 9.8 m/s²
• E (electric field): 4.9 × 10¹⁴ N/C
• q (charge of oil drop): n × e (where 'n' is the number of electrons removed and 'e' is the charge of an electron)
• e (charge of an electron): 1.6 × 10^-19 C

3. Equation:

m \(\times\) g = n \(\times\) e \(\times\) E

4. Solve for 'n':

n = (m \(\times\) g) / (e \(\times\) E)

n = (80 × 10^-9 kg \(\times\) 9.8 m/s²) / (1.6 × 10^-19 C \(\times\) 4.9 × 10¹⁴ N/C)

n = (784 × 10^-9) / (7.84 × 10^-5)

n = 100

Answer: (B) 100

Answer 2

1. Define variables and given information:

• n = number of electrons stripped
• E = 4.9 × 10⁴ N/C (electric field)
• m = 80 μg = 80 × 10⁻⁹ kg (mass of oil drop)
• g = 9.8 m/s² (acceleration due to gravity)
• e = 1.6 × 10⁻¹⁹ C (charge of an electron)

2. Set up the force balance equation:

The electric force (F_e) on the oil drop must balance the gravitational force (F_g) for the drop to be balanced:

\[F_e = F_g\]

3. Express the forces in terms of given quantities:

The electric force is given by:

\[F_e = qE\]

where q is the charge on the oil drop. Since n electrons are stripped, the charge is:

\[q = ne\]

The gravitational force is given by:

\[F_g = mg\]

4. Substitute and solve for n:

Substituting the expressions for the forces into the balance equation:

\[qE = mg\]

\[neE = mg\]

\[n = \frac{mg}{eE}\]

\[n = \frac{(80 \times 10^{-9} \, kg)(9.8 \, m/s^2)}{(1.6 \times 10^{-19} \, C)(4.9 \times 10^4 \, N/C)}\]

\[n = \frac{784 \times 10^{-9}}{7.84 \times 10^{-15}} = 100 \times 10^6 \times 10^{-9} = 100\]

Final Answer: The final answer is \(\boxed{B}\)