Understanding the Principles
* Balance of Forces: The oil drop is suspended, meaning the upward electric force on the charged drop is balanced by the downward gravitational force.
* Electric Force: The electric force on a charged particle in an electric field is given by F = qE, where:
* F is the electric force
* q is the charge on the particle
* E is the electric field strength
* Gravitational Force: The gravitational force on an object is given by F = mg, where:
* F is the gravitational force
* m is the mass of the object
* g is the acceleration due to gravity (approximately 9.8 m/s²)
Steps to Solve
1. Calculate the Volume of the Oil Drop:
* V = (4/3)πr³, where r is the radius (1.64 m)
* V ≈ 1.84 x 10⁻⁵ m³
2. Calculate the Mass of the Oil Drop:
* m = ρV, where ρ is the density (0.851 g/cm³ = 851 kg/m³)
* m ≈ 1.56 x 10⁻² kg
3. Equate the Electric and Gravitational Forces:
* qE = mg
4. Solve for the Charge (q):
* q = (mg) / E
* q ≈ (1.56 x 10⁻² kg * 9.8 m/s²) / (1.92 x 10⁵ N/C)
* q ≈ 8.01 x 10⁻⁸ C
5. Express the Charge in Terms of 'e':
* The elementary charge, 'e', is approximately 1.602 x 10⁻¹⁹ C.
* Divide the calculated charge by the elementary charge:
* q/e ≈ (8.01 x 10⁻⁸ C) / (1.602 x 10⁻¹⁹ C) ≈ 5.00 x 10¹¹
Therefore, the charge on the oil drop is approximately 5.00 x 10¹¹ times the elementary charge 'e'.
Important Note: The answer is extremely large, indicating that the calculated charge is likely not accurate. The oil drop experiment typically deals with charges on the order of a few 'e's. It's possible that there's an error in the given information (the radius, density, or electric field strength).
