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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

Define “fringe field.”

A capacitor made of two parallel uniformly charged circular metal disks carries a charge of +Q and −Q on the inner surfaces of the plates and very small amounts of charge +q and −q on the outer surfaces of the plates. Each plate has a radius R and thickness t, and the gap distance between the plates is s. How much charge q is on the outside surface of the positive disk, in terms of Q?

A large, thin plastic disk with radiusR = 1.5 m carries a uniformly distributed charge of −Q = −3 × 10−5 C as shown in Figure 15.59. A circular piece of aluminum foil is placed d = 3 mm from the disk, parallel to the disk. The foil has a radius of r = 2 cm and a thickness t = 1 mm.


(a) Show the charge distribution on the close-up of the foil. (b) Calculate the magnitude and direction of the electric field at location × at the center of the foil, inside the foil. (c) Calculate the magnitude q of the charge on the left circular face of the foil.

If the total charge on a uniformly charged rod of length is 0.4 m is 2.2 nC, what is the magnitude of the electric field at a location 3 cm from the midpoint of the rod?

Consider a thin glass rod of length L lying along the x axis with one end at the origin. The rod carries a uniformly distributed positive charge Q.

At a location d > L, on the x axis to the right of the rod in Figure 15.56, what is the electric field due to the rod? Follow the standard four steps. (a) Use a diagram to explain how you will cut up the charged rod, and draw the contributed by a representative piece. (b) Express algebraically the contribution each piece makes to the electric field. Be sure to show your integration variable and its origin on your drawing. (c) Write the summation as an integral, and simplify the integral as much as possible. State explicitly the range of your integration variable. Evaluate the integral. (d) Show that your result is reasonable. Apply as many tests as you can think of

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