in a millking oil drop eperiment horizontally place are 0.5cm apart when there is no p.d across the plate the terminal speed of an oil drop is 1.75*10-4-5ms if thr potentail supply is switched on with the upper plate positve the drop is observed to remain stationary when the p.d between the plate is 470.7v
http://www.physics.uci.edu/~advanlab/millikan.pdf
Everything is in the link above. Note E = voltage/distance between plates
To explain how to calculate the charge on an oil drop in a milking oil drop experiment, we need to understand the key concepts involved.
1. Electric Field: The electric field (E) between two charged plates can be calculated using the formula E = V/d, where V is the potential difference (voltage) across the plates, and d is the distance between the plates.
2. Electric Force: The electric force (F) acting on a charged particle is given by the equation F = qE, where q is the charge on the particle and E is the electric field strength.
3. Drag Force: In the milking oil drop experiment, the drag force acting on the oil drop is equal and opposite to the gravitational force acting on it. This drag force is responsible for the terminal velocity of the oil drop and can be calculated using Stoke's Law: F_drag = 6πηrv, where η is the viscosity of air, r is the radius of the oil drop, and v is the velocity of the drop.
Using the given information in your question:
- When there is no potential difference (p.d.) across the plates, the terminal speed of the oil drop is given as 1.75 * 10^(-5) m/s.
- When the potential supply is switched on, and the upper plate is positive, the drop remains stationary at a potential difference of 470.7 V.
To find the charge on the oil drop, we can equate the electrical force on the drop to the gravitational force acting on it when it is stationary.
1. Calculate the electric field between the plates:
E = V/d = 470.7 V / 0.5 cm = 941.4 V/m
2. Calculate the gravitational force acting on the stationary oil drop:
F_gravity = mg, where m is the mass of the drop and g is the acceleration due to gravity (approximately 9.8 m/s^2).
3. Equate the electric force to the gravitational force:
qE = mg
4. Rearrange the equation to solve for the charge (q):
q = mg / E
5. Substitute the values:
q = (mass of the oil drop) * 9.8 / 941.4
Note: To find the mass of the oil drop, we need additional information, such as the density of the oil and the volume of the drop.
By following these steps and plugging in the correct values for the density, volume, and other variables specific to your experiment, you can determine the charge on the oil drop in the milking oil drop experiment.