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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 distance x from the center of the cube, was

14πδo50QLx3

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

Short Answer

Expert verified

Answer

The equation should produce the electric field for a point charge for a large distance of the cube from the center of the electric field, but it cannot produce the equation.

Step by step solution

01

Identification of given data

The given data is listed below as:

  • The edge length of the cube is, L

  • The charge of the cube is, Q

  • The distance of the cube from the center is, x

02

Significance of the magnitude of the electric field

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

03

Determination of the correctness of the equation

The equation of the magnitude of the electric field given in the question is expressed as:

E=14πε050QLx3

Here, 14πε0is the electric field constant, Qis the charge of the cube, Lis the length of the cube and xis the distance of the cube from the center of the electric field.

The equation of the magnitude of the electric field for a point charge is expressed as:

E1=14πε0qr2

Here, 14πε0is the electric field constant, qis the charge of an object and ris the distance of the object from the center of the electric field.

The expression given in the question is wrong as at a certain distance from the cube, the electric field is approximately same as the electric field having a point charge, but the field is not same.

Thus, the equation should produce the electric field for a point charge for a large distance of the cube from the center of the electric field, but it cannot produce the equation.

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Most popular questions from this chapter

Consider a thin plastic rod bent into a semicircular arc of radius Rwith center at the origin (Figure 15.57). The rod carries a uniformly distributed negative charge -Q.

(a) Determine the electric field E→at the origin contributed by the rod. Include carefully labeled diagrams, and be sure to check your result. (b) An ion with charge -2eand mass is placed at rest at the origin. After a very short time ∆tthe ion has moved only a very short distance but has acquired some momentum .P→Calculate P→.

Question: Breakdown field strength for air is roughly . If the electric field is greater than this value, the air becomes a conductor. (a) There is a limit to the amount of charge that you can put on a metal sphere in air. If you slightly exceed this limit, why would breakdown occur, and why would the breakdown occur very near the surface of the sphere, rather than somewhere else? (b) How much excess charge can you put on a metal sphere of radius without causing breakdown in the neighboring air, which would discharge the sphere? (c) How much excess charge can you put on a metal sphere of onlyradius? These results hint at the reason why a highly charged piece of metal tends to spark at places where the radius of curvature is small, or at places where there are sharp points.

A solid metal ball of radius 1.5 cm bearing a charge of −17 nC is located near a solid plastic ball of radius 2 cm bearing a uniformly distributed charge of +7 nC (Figure 15.62) on its outer surface. The distance between the centers of the balls is 9 cm. (a) Show the approximate charge distribution in and on each ball. (b) What is the electric field at the center of the metal ball due only to the charges on the plastic ball? (c) What is the net electric field at the center of the metal ball? (d) What is the electric field at the center of the metal ball due only to the charges on the surface of the metal ball?

A thin rod lies on the x axis with one end atand the other end at-A, as shown in Figure 15.51. A charge of-Q
is spread uniformly over the surface of the rod. We want to set up an integral to find the electric field at location <0,Y,0>due to the rod. Following the procedure discussed in this chapter, we have cut up the rod into small segments, each of which can be considered as a point charge. We have selected a typical piece, shown in red on the diagram

Answer using the variables x,y,dx,A,Qas appropriate. Remember that the rod has charge-Q. (a) In terms of the symbolic quantities given above and on the diagram, what is the charge per unit length of the rod? (b) What is the amount of chargedQon the small piece of lengthdx? (c) What is the vector from this source to the observation location? (d) What is the distance from this source to the observation location? (e) When we set up an integral to find the electric field at the observation location due to the entire rod, what will be the integration variable?

If the magnitude of the electric field in air exceeds roughly 3 × 106 N/C, the air brake down and a spark form. For a two-disk capacitor of radius 47 cm with a gap of 1 mm, what is the maximum charge (plus and minus) that can be placed on the disks without a spark forming (which would permit charge to flow from one disk to the other)?

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