Stu

Dear Anonymous, better than being an a*s like you. You do not know anythimg about me or where I could be coming from. You are a complete a*s. Go eat some hay or something.

I hope your posts make you feel taller! They definatly make me feel happy for my accomplisment and my studying. You keep on telling your self you are tall. Sure you will get there one day when you develop respect for others and what they do as well as for effort they make to reach there goals; and when you don't need to hide from what you say.

A*s:

a hoofed mammal of the horse family, which is typically smaller than a horse and has longer ears and a braying call.

synonyms:donkey; 

From Stu. 

I cannot see why in solution 1 that t+ c falls onto the denominator or why in solution 2 the -1 goes to the numirator. Thank you for the assistance today. Long day for how much i have done. Next i need to resolve null members and null joints. Oh before that i have to work on understanding the integration factor which seems not to hard if i follow the steps. One hopes anyway.

It is easier also.

Oh i was there. Thanks.

What makes you solve it as P= and move everything else?

Question 1 of my last post. I am not sure why it is positive answer from your post. That is that i should or shouldnt use signs when using Fe equations in front of proton charge or electron charge? As well i am unsure what is R^3 in your equation. We have gone through materials but resolved everything to Fe=kq1q2 / r2, F=q0E=kq/r2, v=kq1q2/r. Oh and here is (in my notes) q=Qr^3/R^3 WHICH IS THE DIFFERENCE OF A SHELL IN AN INSULATOR WHERE R IS the outter radius.  

 For question 2 of the avove post. what i did wrong which on posting last night i kind of figured OUT was: firstly I took e to be electron. I took F to be Fe which iM doing a lot. And the rest of my approach was ok but that my calculator is tempremental when solving for tan-1(3.3/6.9) which it cannot do in one step accurately. 

Yes e is in terms of e. I recall now. Still my answer with the same approach was way off. Shocking. 

Here is two more fails. Though I believe my intial answers were close to correct, before trying alternatives values. 

Q1001: A certain atom has 88 protons. Assume that the nucleus is a sphere with radius 7.04 fm and with the charge of the protons uniformly spread through the sphere. At the nucleus surface what are (a) the magnitude and (b) direction (radially inward or outward) of the electric field produced by the protons?

For the above equations I worked out Fe / into 7.04 x by (10^-15)^2 all x by 88. This was wrong, so I / by 88 to get initial Fe per partical which was wrong. 

Q2014: In the figure particle 1 of charge +4e is above a floor by distance d₁ = 3.30 mm and particle 2 of charge +8e is on the floor, at distance d₂ = 6.90 mm horizontally from particle 1. What is the x component of the electrostatic force on particle 2 due to particle 1?

t

Number

Here the solution I found was more complicated perhaps than needed to be. Can you check if it should work? I calulated (tan^-1(o/a)) = 27.40 degrees. Next, Fe = 9E^-9*4*1.6E^-19*8*1.6E^-19. The equation Fecos(27.40) was used to solve the x axis component of force. I think I left our r^2 a few times in here which is 7.64852927E-3. 

{Please elaorate where solving for one partical acting on another is involved and how that is different to using Fe.}

What does this look like graphically?