Question

A drop of oil of mass \(5 \times 10^{-16} \mathrm{~kg}\) is at rest on the bottom plate of parallel plate combination when the electric field is zero. An electric field of \(4 \times 10^3 \mathrm{~V} / \mathrm{m}\) is then applied between the plates, accelerating the drop towards the top plate. What will be the resultant acceleration of the drop if it carries a negative charge of 3 \(\times 10^{-18} \mathrm{C}\) ? (Note: Neglect the effects of air resistance.) 1\. \(9.8 \mathrm{~m} / \mathrm{s}^2\) 2\. \(14.2 \mathrm{~m} / \mathrm{s}^2\) 3\. \(24.0 \mathrm{~m} / \mathrm{s}^2\) 4\. \(\quad 28.4 \mathrm{~m} / \mathrm{s}^2\)

Short answer

We follow the steps to calculate the acceleration of the oil drop. #Step 1: Determine the gravitational force acting on the oil drop: \( F_g = (5 \times 10^{-16}\,\mathrm{kg}) \times 9.8\, \mathrm{m/s^2} = 4.9 \times 10^{-15}\,\mathrm{N} \) #Step 2: Determine the electric force acting on the oil drop: \( F_e = (-3 \times 10^{-18} \,\mathrm{C}) \times (4 \times 10^3 \,\mathrm{V/m}) = -1.2 \times 10^{-14} \,\mathrm{N} \) #Step 3: Calculate the net force on the oil drop: \( F_{net} = F_e - F_g = -1.2 \times 10^{-14}\,\mathrm{N} - 4.9 \times 10^{-15}\,\mathrm{N} = -1.69 \times 10^{-14}\,\mathrm{N} \) #Step 4: Use Newton's second law to find the acceleration of the oil drop: \( a = \frac{F_{net}}{m} = \frac{-1.69 \times 10^{-14}\,\mathrm{N}}{5 \times 10^{-16}\,\mathrm{kg}} = -33.8\,\mathrm{m/s^2} \) Since the charge is negative, the acceleration is actually in the upward direction, making the magnitude of the net acceleration: \( a_{net} = a - g = (-33.8 + 9.8)\,\mathrm{m/s^2} = -24.0\,\mathrm{m/s^2} \) The correct answer is option 3, with a magnitude of \(24.0\,\mathrm{m/s^2}\).

Step-by-step solution

Determine the gravitational force acting on the oil drop

The gravitational force on the oil drop is given by the formula: \( F_g = m \times g \) where \(F_g\) is the gravitational force, \(m\) is the mass of the oil drop, and \(g\) is the acceleration due to gravity (approximately \(9.8~\mathrm{m/s^2}\)). We are given \(m = 5 \times 10^{-16} \mathrm{kg}\), so we can now calculate the gravitational force: \( F_g = (5 \times 10^{-16}\,\mathrm{kg}) \times 9.8 \,\mathrm{m/s^2} \)

Determine the electric force acting on the oil drop

The force acting due to the electric field is given by formula: \( F_e = q \times E \) where \(F_e\) is the electric force, \(q\) is the charge of the oil drop, and \(E\) is the electric field strength. We are given \(q = -3 \times 10^{-18} \,\mathrm{C}\) and \(E = 4 \times 10^3 \,\mathrm{V/m}\), so we can calculate the electric force: \( F_e = (-3 \times 10^{-18} \,\mathrm{C}) \times (4 \times 10^3 \,\mathrm{V/m}) \)

Calculate the net force on the oil drop

Since the electric force acts upward and the gravitational force acts downward, we can subtract the gravitational force from the electric force to find the net force on the oil drop: \( F_{net} = F_e - F_g \)

Use Newton's second law to find the acceleration of the oil drop

Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration: \( F_{net} = m \times a \) where \(a\) is the acceleration of the oil drop. We can solve for the oil drop's acceleration: \( a = \frac{F_{net}}{m} \) Substitute the expressions for \(F_{net}\) and \(m\) and calculate the acceleration, then compare it to the given options to select the right one.

Key concepts

LSAT Prep

Preparing for the Law School Admission Test (LSAT) is a crucial step for aspiring law students. The LSAT is designed to measure skills that are considered essential for success in law school, such as critical reading, analytical reasoning, and logical reasoning.

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Logical Reasoning

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