Static Electricity Legacy Problem #15 Guided Solution
Problem*
The electric field intensity at a particular location surrounding a Van de Graaff generator is 4.5x103 N/C. Determine the magnitude of the force which this field would exert upon …
- An electron when positioned at this location.
- A charged balloon with 1.8 μC of charge when positioned at this location.
- A pith ball with 6.8x10-8 C of charge when positioned at this location.
Audio Guided Solution
This question has to do with an electric field, and the concept of an electric field is that any object which is charged alters the electrical properties of the space which surrounds it in such a manner that any charged object brought into that space will experience an interaction with the space in such a manner that there is a force acting upon that object. Now we measure the strength or intensity of an electric field in terms of the amount of force that acts upon the object brought into the space. That is in Newtons per Coulomb, where the Newtons is the force upon the object brought into the space, and the Coulombs is the quantity of charge of the object that is brought into the space, not of the Van de Graaff generator, but of the object brought into the space. So if you think about that definition and those units, when we say the electric field intensity, E, is equal to 4.5 times 10 to the third, or 4,500 Newtons per Coulomb, we mean that for every one Coulomb of charge brought into that location, there is 4,500 Newtons of force on the object. So we write the equation E equals F divided by Q, where Q is the quantity of charge on the so-called test object brought into the space. And so if we want to find the force acting upon the object, we can rearrange the equation and say force equals E times Q, where E here is 4.5 times 10 to the third, and Q is simply the quantity of charge in part A, B, and C of this problem. Now in part A, that quantity of charge is 1.6 times 10 to the negative 19th charge on an electron, so you need to multiply it by 4.5 times 10 to the third Newtons, and you get 7.2 times 10 to the negative 16th, and that's in units of Newtons. And then in part B, 1.8 mu C is a micro Coulomb, and so that's 1.8 times 10 to the negative 6th Coulombs, you need to multiply that by 4.5 times 10 to the third Newtons per Coulomb, and you would get 8.1 times 10 to the negative third. And finally the pith ball in part C is 6.8 times 10 to the negative 8th Coulombs, and if you multiply that by the value of E, you get 3.06 times 10 to the negative 4th, and you can round that to two significant digits, such as 3.1 times 10 to the negative 4th.
Solution
- 7.2x10-16 N
- 8.1x10-3 N
- 3.1x10-4 N
Habbits of an Effective Problem Solver
- Read the problem carefully and develop a mental picture of the physical situation. If necessary, sketch a simple diagram of the physical situation to help you visualize it.
- Identify the known and unknown quantities; record them in an organized manner. A diagram is a great place to record such information. Equate given values to the symbols used to represent the corresponding quantity - e.g., \(Q_1 = 2.4 \unit{\micro\coulomb}\); \(Q_2 = 3.8 \unit{\micro\coulomb}\); \(d = 1.8 \unit{m}\); \(F_\text{elect} = \colorbox{gray}{Unknown}\).
- Use physics formulas and conceptual reasoning to plot a strategy for solving for the unknown quantity.
- Identify the appropriate formula(s) to use.
- Perform substitutions and algebraic manipulations in order to solve for the unknown quantity.
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