An oil-drop, that has a mass of 3.5 × 10-15 kg and an excess of 7 electrons, is located between horizontal parallel plates. The plates are separated by a distance of 2.5 cm and have and potential diffeDetermine the magnitude and direction of the acceleration on the charged particle.rence of 4.8 × 102 V between them.
F = q E
where
E = 4.8*10^2 / 2.5*10^-2
q = 7 * charge of electron
then
a = F/m
where m = 3.5 * 10^-15 Kg
To determine the magnitude and direction of the acceleration on the charged particle, we can use the equation for the electric force on a charged particle.
1. Calculate the charge of the oil-drop:
The excess of 7 electrons means the oil-drop has a charge of -7e, where e is the charge of an electron (-1.6 × 10^-19 C). So the charge of the oil-drop is:
q = -7e = -7 × (-1.6 × 10^-19 C) = 1.12 × 10^-18 C
2. Calculate the electric field between the parallel plates:
The electric field (E) between the plates can be calculated using the formula:
E = V/d
where V is the potential difference between the plates (4.8 × 10^2 V) and d is the distance between the plates (2.5 cm or 0.025 m).
E = (4.8 × 10^2 V) / (0.025 m) = 1.92 × 10^4 V/m
3. Calculate the electric force on the oil-drop:
The electric force (F) on a charged particle in an electric field is given by the formula:
F = qE
where q is the charge of the oil-drop and E is the electric field between the plates.
F = (1.12 × 10^-18 C) × (1.92 × 10^4 V/m) = 2.15 × 10^-14 N
4. Determine the acceleration of the oil-drop:
Since the electric force (F) is equal to the mass (m) of the oil-drop multiplied by its acceleration (a), we can rearrange the equation to solve for acceleration:
a = F/m
a = (2.15 × 10^-14 N) / (3.5 × 10^-15 kg) = 6.14 m/s^2
The magnitude of the acceleration on the charged oil-drop is 6.14 m/s^2. The direction of the acceleration is towards the positive plate because the oil-drop has a negative charge, and opposite charges attract each other.