Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A cube has sides of lengthL=0.300mm. One corner is at the origin (Fig. E22.6). The nonuniform electric field is given byE=(-5.00N/C×m)x+(3.00N/C×m)zk.(a) Find the electric flux through each of the six cube facesS1,S2,S3,S4,S5andS6. (b) Find the total electric charge inside the cube.

Short Answer

Expert verified

(a)

The electric flux through the first side of the cube face surface is0Nm2/C.
The electric flux through the second side of the cube face surface is0.081Nm2/C
The electric flux through the third side of the cube face surface is0Nm2/C
The electric flux through the fourth side of the cube face surface is0Nm2/C
The electric flux through the fifth side of the cube face surface is-0.135Nm2/C
The electric flux through the sixth side of the cube face surface is0Nm2/C

(b)

The total electric charge inside the cube is-4.781×10-13C.

Step by step solution

01

Define free body diagram and given data

The concept of electric flux, its calculation, and the analogy between the flux of an electric field and that of water. Let us imagine the flow of water with a velocityvin a pipe in a fixed direction, say to the right. If we take the cross-sectional plane of the pipe and consider a small unit area given by ds from that plane, the volumetric flow of the liquid crossing that plane normal to the flow can be given as vds.

02

Simplify the electrical flux

As equation22.8mentions, the electric flux through a spherical surface inside the inner surface of the sphere is given by:

ϕE=Qenclosedε0

As equation 22.5mentions, the electric flux through a surface is given by

localid="1665115999561" ϕE=EAcosf=E×dA

Solve for the first side of the cube faceS1:

ϕS1=E×nS1A=E×^A=0

So, the electric flux through the first side of the cube face surface is0Nm2/C.

Solve for the second side of the cube faceS2:

localid="1665116435683" ϕS2=E×n^S2A=E×k^A=3N/C×m×z×0.3m2=3N/C×m×0.3m×0.3m2=3N/C×m×0.3m3=0.081N×m2/C

So, the electric flux through the second side of the cube face surface is0.081Nm2/C.
Solve for the third side of the cube faceS3:

ϕS3=E×n^3A=E×^A=0N×m2/C

So, the electric flux through the third side of the cube face surface is0Nm2/C

Solve for the fourth side of the cube faceS4:

ϕS4=E×n^S4A=E×k^A=-3N/C×m×z×0.3m2=-3N/C×m×0m×0.3m2=0N×m2/C

So, the electric flux through the fourth side of the cube face surface is0Nm2/C.
Solve for the fifth side of the cube faceS5:

ϕS5=E×n^S5A=E×l^A=-5N/C×m×x×0.3m2=-5N/C×m×0.3m×0.3m2=-5N/C×m×0.3m3=-0.135N×m2/C

So, the electric flux through the fifth side of the cube face surface is -0.135Nm2/C.

Solve for the sixth side of the cube faceS6:

ϕS6=E×S^6A=-E×ι^A=-5N/C×m×x×0.3m2=-5N/C×m×0m×0.3m2=0N×m2/C

So, the electric flux through the sixth side of the cube face surface is0Nm2/C.

03

Solve total electric charge

Solve for the total electric charge inside the cube:
We will take the data (electric flux through each side of the cube) from part (a). In order to evaluate the total electric charge density inside the cube, we use the following relation:

Φtotal=ΦS=ΦS1+ΦS2+ΦS3+ΦS4+ΦS5+ΦS6=0+0.081+0+0+-0.135+0Nm2/C=-0.054Nm2/C

As equation 22.8mentions, the electric flux through a spherical surface inside the inner surface of the sphere is given by:

Φtotal==ΦS=ΦS1+ΦS2+ΦS3+ΦS4+ΦS5+ΦS6=0+0.081+0+0+-0.135+0Nm2/C=-0.054Nm2/C

So, the total electric charge inside the cube is-4.781×10-13C.

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

An 18-gauge copper wire (diameter 1.02 mm) carries a current

with a current density of 3.2×106Am2. The density of free electrons for

copper is8.5×1028electrons per cubic meter. Calculate (a) the current in

the wire and (b) the drift velocity of electrons in the wire.

The heating element of an electric dryer is rated at 4.1 kW when connected to a 240-V line. (a) What is the current in the heating element? Is 12-gauge wire large enough to supply this current? (b) What is the resistance of the dryer’s heating element at its operating temperature? (c) At 11 cents per kWh, how much does it cost per hour to operate the dryer?

In an L-R-C series circuit, what criteria could be used to decide whether the system is over damped or underdamped? For example, could we compare the maximum energy stored during one cycle to the energy dissipated during one cycle? Explain.

You connect a battery, resistor, and capacitor as in Fig. 26.20a, where R = 12.0 Ω and C = 5.00 x 10-6 F. The switch S is closed at t = 0. When the current in the circuit has a magnitude of 3.00 A, the charge on the capacitor is 40.0 x 10-6 C. (a) What is the emf of the battery? (b) At what time t after the switch is closed is the charge on the capacitor equal to 40.0 x 10-6 C? (c) When the current has magnitude 3.00 A, at what rate is energy being (i) stored in the capacitor, (ii) supplied by the battery

A particle of mass 0.195 g carries a charge of-2.50×10-8C. The particle is given an initial horizontal velocity that is due north and has magnitude4.00×104m/s. What are the magnitude and direction of the minimum magnetic field that will keepthe particle moving in the earth’s gravitational field in the samehorizontal, northward direction?

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