ZuoTi.Pro
Question

Hi,

I'm having trouble solving this problem... Could you please solve it using clear detailed steps so that I can better understand how to tackle it next time?

Thanks! :)

A charge of 100 AHC is at the center of a cube of edge 75.0 cm. No other charges are nearby. (a) Find the total flux through each face of the cube. (b) Find the flux through the whole surface of the cube N m2/c (a) would change O (b) would change (c) Would your answers to parts (a) or (b) change if the charge were not at the center? (Select all that apply.) neither would change Explain

science   physics   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Answer #1

coRDING TO GauSss Lau, THE ToTAL fLUx THROUGH ANY closeD suRFACE CALN WRITTEN AS - E LE CTRI FROM SyMMETRY, SINCE CHARGE·IS THE CENTER oF THE cusE, ALL THE OF THE CUBE GETS ONE xTH OF THE TOTAL FLUx L FLUx, wHicH wILL 81E FACE 6 6 6.85 xio-12 FROM GAUSSs LAh, THE TOTAL ELECTRIC FLUX THROUGH WHoLE SURFACE OF CUBE IS - QV 8.85 x10-12 CHARGE IS NOT AT THE CENTER OF THE CUBE , TH E TOTAL FLUX REMAINS SAME, SINCE ENCLOSED CHARGE REMAINS SAME GUT NOW IN THE A6SENCE OF SYMMETRY, EvERY FACE MAY GET DIFFERENT AMDUNT OF FLUX THus_ OPTION! (a) | IS CORRECT.

0 0
Add a comment
Know the answer?
Add Answer to:
Hi, I'm having trouble solving this problem... Could you please solve it using clear detailed steps...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions

  • Hi, I'm having trouble solving this problem... Could you please solve it using clear detailed steps...

    Hi, I'm having trouble solving this problem... Could you please solve it using clear detailed steps so that I can better understand how to tackle it next time? Thanks! :) Consider the following image. (Take q1 +2.95 nC and q28.44 nC.) gh 42 (a) Find the net electric flux through the cube shown in the figure above. N m2/c (b) Can you use Gauss's law to find the electric field on the surface of this cube? Yes No Explain

  • Hi, I'm having trouble solving this problem... Could you please solve it using clear detailed steps...

    Hi, I'm having trouble solving this problem... Could you please solve it using clear detailed steps so that I can better understand how to tackle it next time? Thanks! :) A particle with a charge of -60.0 nC is placed at the center of a nonconducting spherical shell of inner radius 20.0 cm and outer radius 31.0 cm. The spherical shell carries charge with a uniform density of-3.88 μC/m3. A proton moves in a circular orbit just outside the spherical...

  • 11. 10 points SerCP11 15.8.P.052 A charge of 1.87 x 102 uC is at the center...

    11. 10 points SerCP11 15.8.P.052 A charge of 1.87 x 102 uC is at the center of a cube of edge 63.0 cm. No other charges are nearby. (a) Find the flux through the whole surface of the cube. N m2/C (b) Find the flux through each face of the cube. N m2yc (c) Would your answers to parts (a) or (b) change if the charge were not at the center? (Select all that apply.) (a) would change (b) would...

  • A charge of 1.60 ✕ 10^2 μC is at the center of a cube of edge...

    A charge of 1.60 ✕ 10^2 μC is at the center of a cube of edge 89.0 cm. No other charges are nearby. (a) Find the flux through the whole surface of the cube. N · m^2/C (b) Find the flux through each face of the cube. N · m^2/C (c) Would your answers to parts (a) or (b) change if the charge were not at the center? (Select all that apply.) (a) would change (b) would change neither would...

  • A Charge Of 1.30 HW-2 Chapter 15.3-5, 15.8 ณ10 points Preveus Answers SerCP11 158P052. A charge...

    A Charge Of 1.30 HW-2 Chapter 15.3-5, 15.8 ณ10 points Preveus Answers SerCP11 158P052. A charge of 1.42 x 10 pC is at the center of a cube of edge 80.5 cm. No other charges are nearby. (a) Find the flux through the whole surface of the cube Your response differs from the correct answer by more than 10%. Double check your calculations. N . mic (b) Find the flux through each face of the cube N m2/c answers to...

  • A charge of 200 µC is at the center of a cube of edge 65.0 cm....

    A charge of 200 µC is at the center of a cube of edge 65.0 cm. No other charges are nearby. (a) Find the total flux through each face of the cube.    Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. N·m2/C (b) Find the flux through the whole surface of the cube. N·m2/C (c) Would your answers to parts (a) or (b) change if the charge were not...

  • A charge of 155 µC is at the center of a cube of edge 95.0 cm....

    A charge of 155 µC is at the center of a cube of edge 95.0 cm. No other charges are nearby. (a) Find the flux through each face of the cube. N · m2/C (b) Find the flux through the whole surface of the cube.N · m2/C

  • 8, AGA charge of 170 μC is at the center of a cube of edge 80.0...

    8, AGA charge of 170 μC is at the center of a cube of edge 80.0 cm. No other charges are nearby. (a) Find the flux a/c ni- ric be through each face of the cube. (b) Find the flux through the whole surface of the cube. (c) What If? Would your answers to either part (a) or part (b) change if the charge were not at the center? Explain. ind inside a submarine:

  • I'm having some trouble with this question. Not sure how to go about solving it. Any...

    I'm having some trouble with this question. Not sure how to go about solving it. Any help would be appreciated. Thanks. An object is made up of a disk (M=7.0kg, R=2.0m) and two point masses (m1=3.0kg and m2=4.0kg). The arrangement of the disk and the point masses are shown in the figure and the dimensions are d1=1.0m, d2=1.5m, d3=0.5m, d4=1.0m, and t=0.3m. A. What is the x-coordinate of the center of mass of the object? B. What is the z-coordinate...

  • Hi. I'm having trouble solving the rate law for my chemistry lab report sheet. I'm unsure...

    Hi. I'm having trouble solving the rate law for my chemistry lab report sheet. I'm unsure if I'm doing it right. Below is my trials To calculate rate law, I choose experiment 1 and 2 for [I]x and experiment 2 and 3 for [H2O2]y. I got x=1 and y=1. This would make my rate law: rate = k [I-] [H2O2] Is this correct? Please help! Thanks in advance Trial [1] [H202] Rate (M/s) 1 3.00 x 10^-2 M 1.50 x...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT