Would the self-created electric field at the end of a pointed conductor, such as a lightning rod, remove positive or negative charge from the conductor? Would the same sign charge be removed from a neutral pointed conductor by the application of a similar externally created electric field? (The answers to both questions have implications for charge transfer utilizing points.)

Short Answer

Expert verified

The self-created electric field on an uneven conductor would remove positive charges, if the field is created due to positive charges and it would remove negative charges if the field is created due to negative charges. The externally applied field will remove the opposite kind of charge from the conductor.

Step by step solution

01

Distribution of charges

When the charge is given to a conductor of uneven shape, the charge tends to concentrate at the sharp edges, giving rise to a strong electric field at the edges compared to its body.

02

Electric field due to positive charge

If a positive charge is given to a conductor of uneven shape, the positive charge concentrates at the sharp edge of the conductor where the curvature of the surface is small. When the concentration of charges at the sharp edges is high, this will result in the creation of a large electric field at the edges.

Sometimes the electric field is strong enough that it would ionize the molecules of air nearer to it. The negative charges from the ionized air molecules would flow towards the positively charged sharp edges and neutralize them.

Hence, the positive charges from the sharp edge will be removed.

03

Electric field due to the negative charge

In a similar manner, if the object is given a negative charge, the electric field due to the negative charge will attract the positive charge from the ionized air.

Hence, the negative charge from the sharp edge will be removed.

04

External electric field

If an electric field created by a positive charge is applied to a neutral conductor, the field would induce negative charges on the sharp edges of the conductor. These negative charges would then be removed when they come in contact with air molecules.

In a similar manner, if the field is created by negative charges, the field will induce positive charges on the sharp edge of the conductor. These positive charges would then be removed when they come in contact with air molecules.

Thus, the externally applied field would remove the opposite charge from the conductor and the self-created field would remove a similar type of charge from sharp edges.

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Most popular questions from this chapter

Point charges of\(25.0{\rm{ }}\mu {\rm{C}}\)and\(45.0{\rm{ }}\mu {\rm{C}}\)are placed\(0.500{\rm{ m}}\)apart. (a) At what point along the line between them is the electric field zero? (b) What is the electric field halfway between them?

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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 electron’s original horizontal velocity. (These can be used to change the electron’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.

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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

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