+1313 Hi, I am not sure if it's input error or not for my caluculations. I resolve Fe=kq1q2/r2 for a. I used q1/e for b. Can you solve for me. No try's left on the test if you had concerns.
Two tiny, spherical water drops, with identical charges of -9.07 × 10-16 C, have a center-to-center separation of 1.43 cm. (a) What is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?
| (a) | 0.0000076 |  | Units | This answer has no units ° (degrees) m kg s m/s m/s^2 N J W N/m kg·m/s or N·s N/m^2 or Pa kg/m^3 m/s^3 times |
| (b) | 5.67E-35 | |
+1313 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.}
+33661 Q1001.
The following image should help here:
Q will be 88*|e|.
Q2014.
Use the standard equation for force, F, (as in my first post above) with q1 = 4|e| and q2 = 8|e|. The distance, r, is √(3.32 + 6.92) mm (You might want to change this into metres to make sure your units are consistent).
The angle of interest, θ, is tan-1(3.3/6.9) (which I get to be 25.56°, not 27.4°), giving the component of force along the x-axis as F*cos(θ).
NB k ≈ 9E+9 not 9E-9.
The particles act on each other with the same magnitude force (opposite directions):- Newton's 3rd law of motion.
+1313 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.
+33661 "Question 1 of my last post. I am not sure why it is positive answer from your post."
The field direction is positive, by convention, going from positive to negative, so it must be going away from the positives spread through the sphere.
"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"
Use negative sign when it's an electron and positive when a proton. R is the radius of the sphere (have a look at the sphere drawn just below the graph), given as 7.04fm (7.04 femto metres = 7.04x10-15m in your question), R^3 is R raised to the power 3 (or R3).
