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Chapter 21: Problem 7

An average human weighs about 650 N. If each of two average humans could carry 1.0 C of excess charge, one positive and one negative, how far apart would they have to be for the electric attraction between them to equal their 650-N weight?

Short Answer

Expert verified
The humans need to be approximately 3720 meters apart.

Step by step solution

01

Understand the Electric Force Formula

The electric force between two point charges is given by Coulomb's Law: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \(F\) is the force between the charges, \(k\) is Coulomb's constant (\(8.99 \times 10^9 \text{ N m}^2/\text{C}^2\)), \(q_1\) and \(q_2\) are the charges and \(r\) is the distance between them.
02

Set the Known Values

Given that \(q_1 = q_2 = 1.0 \text{ C}\) and we want the electric force \(F\) to equal 650 N, we have:- \(q_1 = 1.0 \text{ C}\)- \(q_2 = 1.0 \text{ C}\)- \(F = 650 \text{ N}\)
03

Plug in the Values into Coulomb's Law

Inserting the known values into Coulomb's Law gives:\[ 650 = (8.99 \times 10^9) \frac{(1.0)(1.0)}{r^2} \]
04

Solve for Distance \(r\)

Rearrange the equation to solve for \(r\):\[ r^2 = \frac{(8.99 \times 10^9)}{650} \]Calculate \(r^2\):\[ r^2 = \frac{8.99 \times 10^9}{650} = 1.38307 \times 10^7 \]\[ r = \sqrt{1.38307 \times 10^7} \approx 3720 \text{ m} \]
05

Conclusion

The two humans with charges would need to be approximately 3720 meters apart for the electric attraction to equal their 650 N weight.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Force
Understanding the concept of electric force is essential to grasp Coulomb's Law effectively. Electric force is the force of attraction or repulsion between two electrically charged objects, also known as point charges. According to Coulomb's Law, this force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. This implies that as the charges increase, the force between them becomes stronger. Conversely, as the distance between the charges increases, the force decreases.

Coulomb's Law can be mathematically represented as: \[ F = k \frac{|q_1 q_2|}{r^2} \] where
  • \(F\) is the electric force
  • \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \text{ N m}^2/\text{C}^2\)
  • \(q_1\) and \(q_2\) are the magnitudes of the charges
  • \(r\) is the distance between the charges
Coulomb's Law highlights the nature of electric forces, which can be either attractive or repulsive depending on the types of the charges involved. Like charges repel each other while unlike charges attract.
Point Charges
The term 'point charges' refers to charged objects that are considered to be infinitesimally small. In the context of our exercise, understanding point charges is crucial as they simplify the calculation of electric force. By approximating real-world objects as point charges, we ignore the specifics of their shape and size, focusing only on the amount of charge they possess.

Point charges are useful in simplifying problems and analyzing electric fields and forces between objects without the complexity of their physical structure. When dealing with point charges, the electric force calculations are precise since the charges occupy a negligible space, allowing the formula \( F = k \frac{|q_1 q_2|}{r^2}\) to be used effectively. Consequently, in many physics problems, objects are treated as point charges when their actual physical dimensions do not significantly affect the calculations.
Distance Calculation
Distance calculation is a vital part of applying Coulomb's Law to determine the electric force between charges. In our exercise, we calculated the required distance by rearranging Coulomb's Law. When the force \(F\) is known, and the charges \(q_1\) and \(q_2\) are given, we solve for distance \(r\) as follows:First, we rearrange the equation: \[ r^2 = \frac{k |q_1 q_2|}{F} \]Then, solve for \(r\): \[ r = \sqrt{\frac{k |q_1 q_2|}{F}} \]In this exercise, both charges were 1.0 C, and we wanted the electric force to equate to a weight force of 650 N. Inserting these into the formula:
  • Calculate \( r^2 = \frac{8.99 \times 10^9}{650} \)
  • Then, \( r = \sqrt{1.38307 \times 10^7} \approx 3720 \text{ meters} \)
Calculating the distance in this manner demonstrates how electric force is inversely related to the square of the distance between the charges. As the distance increases, the force diminishes, illustrating the necessity for significant separation to balance a large force with large charges.

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Most popular questions from this chapter

A proton is placed in a uniform electric field of 2.75 \(\times 10^3 \space N/C\). Calculate (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after 1.00 \(\mu\)s in the field, assuming it starts from rest.

Two point charges are located on the \(y\)-axis as follows: charge \(q_1 = -1.50 \)nC at \(y = -\)0.600 m, and charge \(q_2 = +\)3.20 nC at the origin \((y = 0)\). What is the total force (magnitude and direction) exerted by these two charges on a third charge \(q_3 = +\)5.00 nC located at \(y = -\)0.400 m ?

In a follow-up experiment, a charge of \(+40\) pC was placed at the center of an artificial flower at the end of a 30-cm long stem. Bees were observed to approach no closer than 15 cm from the center of this flower before they flew away. This observation suggests that the smallest external electric field to which bees may be sensitive is closest to which of these values? (a) \(2.4 \space N/C; (b) 16 \space N/C; (c) 2.7 \times 10^{-10} \space N/C; (d) 4.8 \times 10^{-10} \space N/C.\)

Point charges \(q_1 = -\)4.5 nC and \(q_2 = +\)4.5 nC are separated by 3.1 mm, forming an electric dipole. (a) Find the electric dipole moment (magnitude and direction). (b) The charges are in a uniform electric field whose direction makes an angle of 36.9\(^\circ\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.2 \times 10^{-9}\) N \(\cdot\) m ?

(a) An electron is moving east in a uniform electric field of 1.50 N\(/\)C directed to the west. At point \(A\), the velocity of the electron is 4.50 \(\times 10^5\) m\(/\)s toward the east. What is the speed of the electron when it reaches point B, 0.375 m east of point A? (b) A proton is moving in the uniform electric field of part (a). At point \(A\), the velocity of the proton is 1.90 \(\times 10^4\) m\(/\)s, east. What is the speed of the proton at point \(B\)?
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