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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Define “fringe field.”

Consider a capacitor made of two rectangular metal plates of length L and width W, with a very small gap s between the plates. There is a charge +Qon one plate and a charge −Qon the other. Assume that the electric field is nearly uniform throughout the gap region and negligibly small outside. Calculate the attractive force that one plate exerts on the other. Remember that one of the plates doesn’t exert a net force on itself

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?

Explain qualitatively how it is possible for the electric field at locations near the center of a uniformly charged disk not to vary with distance away from the disk.

If the total charge on a thin rod of length 0.4is2.5nC, what is the magnitude of the electric field at a location1Cmfrom the midpoint of the rod, perpendicular to the rod?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free