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Chapter 22, Problem 009DJ Your answer partially correct, Try again.Thc fiqurc shows Y=7.oo m?charged macicles OnanJs;-1.60Cat? - 2.0d160Cal >What are the (2) * a...

Question

Chapter 22, Problem 009DJ Your answer partially correct, Try again.Thc fiqurc shows Y=7.oo m?charged macicles OnanJs;-1.60Cat? - 2.0d160Cal >What are the (2) * and (D) y- components of the nct clectrk ficld produced point _(a) Numbe 000000000255Units NC o Vim:(6) NumberUnits NC Or Vma

Chapter 22, Problem 009 DJ Your answer partially correct, Try again. Thc fiqurc shows Y=7.oo m? charged macicles Onan Js; -1.60 Cat? - 2.0d 160 Cal > What are the (2) * and (D) y- components of the nct clectrk ficld produced point _ (a) Numbe 000000000255 Units NC o Vim: (6) Number Units NC Or Vma



Answers

Complete the table. MISSED THIS? Read Section 1.6; Watch $\mathrm{KCV} 1.6$ a. $1245 \mathrm{~kg} \quad 1.245 \times 10^{6} \mathrm{~g} \quad 1.245 \times 10^{9} \mathrm{mg}$ b. $515 \mathrm{~km}$ _____ dm _____cm c. $122.355 \mathrm{~s}$ _____ms ____ks d. $3.345 \mathrm{~kJ}$ _____J ____mJ

For this problem on the topic of electric potential, we have two point charges each with the magnitude to micro columns that are located on the X axis. One is that X is equal to one m and the other X is equal to minus one m. We want to find the electric potential on the Y axis at Y. Is equal to 0.5 m. We then want to calculate the electric potential energy of a third charge of magnitude minus the micro columns. It is placed on the Y axis at Y is equal to 0.5 m. So firstly the potential V at our point label, P is simply the potential due to charge Q. One which is K E Q one over our one must be potential due to charge too, which is K E E Q two over our two. And this is equal to two into K E Q. Of our. And since the charges that is equal between the two as well as their separation. And so this is to into eight 0.99 times 10 to the nine that's K. Newton meters squared, McCullum squared, multiplied by the magnitude of the charge, which is to times 10 to the minus six columns, all divided by their distance to point P, which is the square root of one meter squared plus zero point five m squared. And so calculating, we get the potential any point that is 0.5 m along the y axis to be three 0.22 times 10 to the four vaults, which is 32 going to kilovolts. Next, we want to find the electric potential energy of a third charge that is placed on the y axis at Y. Is equal to 0.5 m. So the electric potential energy, you is able to cue times V. This is equal to the charge minus three times 10 to the minus six columns, plus the potential there, which is three. What time's the potential? Rather, the 0.22 times 10 to the power four jules, A colon. This gives the electric potential energy at this point to be minus nine 0.65 times 10 to the minus two jules.

So for this question, it is just asking how you use your resource is for this particular chapter. So did you have sufficient time to study? Um, if not, what could you have done to create more time for studying? And you need to answer yes or no? Did you work out all of the assigned homework problems in this chapter? Yes or no? If you encounter difficulty in this chapter, did you see your instructor or to Dover help? Yes or no. And have you taken advantage of the textbook supplements? Yes or no? So just answer honestly, and hopefully can use this for the next chapter to see how you can improve with studying.

For this problem. On the topic of castles law, we have shown cross sections through to large parallel on conducting sheets that have identical distributions of positive charge And self discharge densities of 1.77 times 10 to the -22 columns per square meter. We want to find the electric field points that are above the sheets between the sheets and then below them. Now we know that the electric field due to either sheet of charge with surface charge density, sigma is perpendicular to the plane sheet of to the plane of the sheet and points away from the sheet of the charges positive and has a magnitude E. Is equal to sigma over to absolutely not. And so if we use the superposition principle, we have for both the sheets above them, he is equal to sigma over. Absolutely not, which is 1.77 times 10 To the -22 columns per square meter divided by 8.8, 5 Times 10 to the -12. Cool. Um squared the newton metre squared, which gives us the electric field strength to be too times 10 To the -11. Newton's McCullum pointing in the upper direction, which means that the vector field E is equal to two Times 10 to the -11 newtons per column Jihad. Now, at points Between the two sheets, we know that the electric field must be zero and at points below the sheets. Again, we have the equal to sigma over. Absolutely not using superposition, but this time points downward. And so the electric field vector is minus two Times 10 to the -11 newtons per column jihad.

For port A. The potential at the point of diagram V equals K one Q one over our one plus Que two que two over our two citified. Since about charges are symmetric and equal. Two times K comes constant times. Q. Over our distances separation caring is symmetrical, so we plug in the values V Eagles to times Combs Constant, eight point 99 times 10 tonight Newton We're squared Proclaim square times the charges, which is 2.0 times 10 26 Coombs tend to negative. Six cools Small charges. That's your new rail. Then you divided by for persecuted therm reference referencing a diagram sort of distances. Separation for each one will be a square root one meter squared plus 0.5 readers Square adam up square Root it and then you get a potential I see equals three point to to times 10 26 faults that is party part B. Potential energy equals Q tons of voltage Q. B. So it plug in the values in negative three times 10 to make of six. Coombs multiplied by the potential that we found three point to to times 10 to 6 vaults or Jules per cool, and that gives us a value of negatives. Nine point 65 times 10 10 A second Jules for the potential energy.


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