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)
- 1
- 100
- 10
- 1000
- 10000
The Correct Option is B
Approach Solution - 1
1. Forces involved:
- Gravity (Fg): Pulling downwards. Fg = m \(\times\) g
- Electric Force (Fe): Pushing upwards. Fe = q \(\times\) E
Since the forces are balanced, Fg = Fe.
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
Approach Solution -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 (Fe) on the oil drop must balance the gravitational force (Fg) 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 \(\boxed{B}\)
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Concepts Used:
Electric Field
Electric Field is the electric force experienced by a unit charge.
The electric force is calculated using the coulomb's law, whose formula is:
\(F=k\dfrac{|q_{1}q_{2}|}{r^{2}}\)
While substituting q2 as 1, electric field becomes:
\(E=k\dfrac{|q_{1}|}{r^{2}}\)
SI unit of Electric Field is V/m (Volt per meter).