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Explain briefly how knowing the electric field of a ring helps in calculating the field of a disk.

Short Answer

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Answer

The disk can be divided into tiny rings and the electric fields of the tiny rings can be added to get the electric field of the desk.

Step by step solution

01

Significance of an electric field

The electric field is referred to as a region that helps an electrically charged particle to exert force on another particle. The magnitude of the electric field is directly proportional to the charge of an object and inversely proportional to the distance of that object from the center of the electric field.

02

Determination of the calculation of the electric field of a desk

The magnitude of the electric field of a particular disk can be determined with the help of dividing the disk into rings having tiny thickness. Hence, with the help of that, the electric field of one ring can be calculated and the magnitude of the electric fields of the rings can be added in order to get the electric field of a desk.

Thus, the disk can be divided into tiny rings and the electric fields of the tiny rings can be added to get the electric field of the desk.

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Most popular questions from this chapter

Question: A hollow ball of radius , made of very thin glass, is rubbed all over with a silk cloth and acquires a negative charge of that is uniformly distributed all over its surface. Location A in Figure 15.64 is inside the sphere, from the surface. Location B in Figure 15.64 is outside the sphere, from the surface. There are no other charged objects nearby.


Which of the following statements about , the magnitude of the electric field due to the ball, are correct? Select all that apply. (a) At location A, is . (b) All of the charges on the surface of the sphere contribute to at location A. (c) A hydrogen atom at location A would polarize because it is close to the negative charges on the surface of the sphere. What is at location B?

Define 鈥渇ringe field.鈥

If the total charge on a thin rod of length0.4mis 2.5n/C, what is the magnitude of the electric field at a location1Cmfrom the midpoint of the rod, perpendicular to the rod?

A student claimed that the equation for the electric field outside a cube of edge length L, carrying a uniformly distributed charge Q, at a distancex from the center of the cube, was

role="math" localid="1668495301957" E=Q0Lx1/2

Explain how you know that this cannot be the right equation.

Two rings of radius 5cmare 20cmapart and concentric with a common horizontal axis. The ring on the left carries a uniformly distributed charge of +35nC, and the ring on the right carries a uniformly distributed charge of -35nC. (a) What are the magnitude and direction of the electric field on the axis, halfway between the two rings? (b) If a charge of-5nCwere placed midway between the rings, what would be the magnitude and direction of the force exerted on this charge by the rings? (c) What are the magnitude and direction of the electric field midway between the rings if both rings carry a charge of +35nC?

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