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Why do most objects tend to contain nearly equal numbers of positive and negative charges?

Short Answer

Expert verified

Most objects tend to contain nearly equal numbers of positive and negative charges because nature works on the principle of energy minimization.

Step by step solution

01

Electric charge

Electric charge is a physical or fundamental property of matter due to which it experiences some force when placed near another charged particle.

Electric charge is a scalar quantity and measured in terms of Coulombs.

02

Objects contain nearly equal numbers of positively and negative charges

Nature works on the principle of energy minimization, which means each and every object in the universe tries to attain minimum energy.

Carrying any charge means the existence of force that would require energy, therefore most of the objects tend to contain nearly equal numbers of positive and negative charges so the net charge on the object will be zero. Thus, there will be no non-zero forces and hence energy is minimized.

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Most popular questions from this chapter

What is the force on the charge located at \(x = 8.00{\rm{ }}cm\) in Figure 18.52(a) given that \(q = 1.00{\rm{ }}\mu C\)?

Figure 18.52 (a) Point charges located at \[{\bf{3}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{11}}.{\bf{0}}{\rm{ }}{\bf{cm}}\] along the x-axis. (b) Point charges located at \[{\bf{1}}.{\bf{00}},{\rm{ }}{\bf{5}}.{\bf{00}},{\rm{ }}{\bf{8}}.{\bf{00}},{\rm{ }}{\bf{and}}{\rm{ }}{\bf{14}}.{\bf{0}}{\rm{ }}{\bf{cm}}\] along the x-axis

An electron has an initial velocity of\(5.00 \times {10^6}{\rm{ m}}/{\rm{s}}\)in a uniform\(2.00 \times {10^5}{\rm{ N}}/{\rm{C}}\)strength electric field. The field accelerates the electron in the direction opposite to its initial velocity. (a) What is the direction of the electric field? (b) How far does the electron travel before coming to rest? (c) How long does it take the electron to come to rest? (d) What is the electronโ€™s velocity when it returns to its starting point?

Figure 18.57 shows an electron passing between two charged metal plates that create an\(100{\rm{ N}}/{\rm{C}}\)vertical electric field perpendicular to the electronโ€™s original horizontal velocity. (These can be used to change the electronโ€™s direction, such as in an oscilloscope.) The initial speed of the electron is\(3.00 \times {10^6}{\rm{ m}}/{\rm{s}}\), and the horizontal distance it travels in the uniform field is\(4.00{\rm{ cm}}\). (a) What is its vertical deflection? (b) What is the vertical component of its final velocity? (c) At what angle does it exit? Neglect any edge effects.

Consider identical spherical conducting space ships in deep space where gravitational fields from other bodies are negligible compared to the gravitational attraction between the ships. Construct a problem in which you place identical excess charges on the space ships to exactly counter their gravitational attraction. Calculate the amount of excess charge needed. Examine whether that charge depends on the distance between the centers of the ships, the masses of the ships, or any other factors. Discuss whether this would be an easy, difficult, or even impossible thing to do in practice.

Considering Figure, suppose that qa=qd and qb=qc . First show that q is in static equilibrium. (You may neglect the gravitational force.) Then discuss whether the equilibrium is stable or unstable, noting that this may depend on the signs of the charges and the direction of displacement of qfrom the center of the square.

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