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24. Electric Force & Field; Gauss' Law
Electric Fields in Capacitors
5:05 minutes
Problem 37
Textbook Question
Textbook QuestionA 0.80-μm-diameter oil droplet is observed between two parallel electrodes spaced 11 mm apart. The droplet hangs motionless if the upper electrode is 20 V more positive than the lower electrode. The density of the oil is 885 kg/m³ . (c) Does the droplet have a surplus or a deficit of electrons? How many?
Verified step by step guidance
1
First, calculate the weight of the oil droplet using the formula for weight, W = mg, where m is the mass of the droplet and g is the acceleration due to gravity (approximately 9.8 m/s²). To find the mass, use the volume of the droplet (volume of a sphere = \(\frac{4}{3}\pi r^3\)) and the density given (\(\rho = 885 \text{ kg/m}^3\)).
Next, determine the electric force required to balance the weight of the droplet. Since the droplet is motionless, the electric force (F_e) must equal the gravitational force (W). Use the formula for electric force, F_e = qE, where q is the charge on the droplet and E is the electric field between the electrodes.
Calculate the electric field (E) between the electrodes using the formula E = V/d, where V is the voltage difference between the electrodes and d is the distance between them.
Solve for the charge q on the droplet using the equation F_e = qE = W. Rearrange to find q = W/E.
Determine if the droplet has a surplus or deficit of electrons by considering the sign of the charge q. If q is positive, the droplet has a deficit of electrons (missing electrons); if q is negative, it has a surplus of electrons (extra electrons). Calculate the number of electrons by dividing the magnitude of q by the elementary charge (approximately \(1.6 \times 10^{-19}\) Coulombs).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Electric Force
The electric force is the interaction between charged objects, described by Coulomb's law. In this scenario, the electric field created by the voltage difference between the electrodes exerts a force on the charged oil droplet. This force can either lift or lower the droplet depending on its charge, allowing it to remain suspended when balanced by gravitational force.
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Gravitational Force
Gravitational force is the attractive force between two masses, calculated using Newton's law of universal gravitation. For the oil droplet, this force acts downward due to its mass and the acceleration due to gravity. The balance between this gravitational force and the electric force determines whether the droplet remains motionless between the electrodes.
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Charge and Electron Deficit/Surplus
Charge refers to the property of matter that causes it to experience a force in an electric field. An object with a surplus of electrons is negatively charged, while a deficit indicates a positive charge. In this case, determining whether the droplet has a surplus or deficit of electrons involves analyzing the forces acting on it and calculating the net charge required to balance the gravitational force with the electric force.
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