Question

Hello, I'm taking College Physics. I've been working on practice exam for the past 3 days. I have attached my work and the practice exam. I" having difficulty with problems 14-19 =(. I think that is minor details, if someone could review 1 or 2 I'll be more than thankful.
Thank you!

Answer 1

can we get the answers right side up please?

Answer 2

yes, will do

Answer 3

For 14, please look at 14 Attempt#2

Answer 4

it's stil upside-down

if not, i would link it

sort that out, mate

Answer 5

http://assets.openstudy.com/updates/attachments/57e846ede4b0e5e04a848951-issy14-1474842446769-practiceexam1answers.pdf

Answer 6

Is the link working now?

Answer 7

the links work, but the pages are upside down. You have them scanned the wrong way. You have to rotate the scanned pages images before creating the pdf.

Answer 8

hahaha ok, will do. Hmm so weird.

Answer 9

Here it is, disregard 1 - 8. Those I'm ok with its 14 - 19 that I"m stuck in

Answer 10

page 1 - 3 are tilted but those are irrelevant.

Answer 11

So start from the top!!

Answer 12

just posted Qu 19

Answer 13

question 19, posted where?

Answer 14

new thread!

Answer 15

look up

Answer 16

i'm in new in the website, go it..

Answer 17

As far as 14 goes it looks about right. You are only asked the magnitude of the force so I don't quite get why you bothered with the vector account.
Given that all 3 charges lie on an circular arc with the test charge in the center of that circle it's easy to conclude all charges are equally far away from the test charge. All you had to do is add them algebraically as you did.

As far as the second go at 14, where you employed a bit of vector calculus I've 2 things:
First off I think you missed the cosine and sine positions. Sine is the opposite/hypotenuse in a right angled triangle and cosine is the adjacent/hypotenuse.

So the x component which is the left-right horizontal line is the opposite side of that triangle, the y component of that force is the adjacent (vertical/ up-down) /hypotenuse. So you should rotate those around so that it's sin(theta)i + cos(theta)j
Second thing has to deal with the conversion of 45 degrees to radians, I'm not sure your calcs will be right if you put 45 in, you need to convert that to pi/4.

Answer 18

Coulomb's law, superposition of forces, resolution of vector forces, number crunching ... I don't think that the thing is even in EQUILIBRIUM !
Bon voyage
http://perendis.webs.ccom

Answer 19

Enclosed is my attempt at No 19. I think it's about GAUSS'S law which covers Estatics and Gravitation. BUT my answer isn't in the list.
Maybe someone could check MY logic ...

Answer 20

And again, now that I can see the wood for the trees. The radius being asked for is OUTSIDE the expletive sphere. So, the total charge enclosed is that due to the sphere, and the flux is calculated at the 2R point. I think. E0=E1, which is C isn't it ?

Answer 21

I'm not following how you got the expressions for the electrical field. The integration procedure itself isn't shown but just the algebra that follows after it, so I'm a bit confused. Gauss:
$$\oint_S\vec E\cdot d\vec A = \Phi = \frac{Q_{enc}}{\epsilon_0}$$
Because our Gaussian surface is a shell the dot product between electric field and infinitesimal surface element will be simplified to just EdA in non vector form. This is because the electrical field will be perpendicular to the surface of the shell at all points, which is precisely where the infinitesimal area vector points to - after all it's defined as the normal to the surface with magnitude of the surface.
$$\oint_S Ed A = \frac{Q(V) }{\epsilon_0}\\
E\oint_S dA = \frac{Q(V) }{\epsilon_0}\\
E4\pi r^2 = \frac{ Q(V)}{\epsilon_0}$$
So obviously the only trick here is to be able to figure out what Q did we encircle. For the case when r>R we have the standard point charge because we obviously encircle the *entire* charge Q that is in the sphere.
$$E = \frac{Q}{4\pi\epsilon_0r^2}$$
To figure out how much charge we encircle when r<R, we're going to define the charge volume density, rho, as an average charge per unit volume of the sphere. This makes sense because you suppose the charges are repelling and spacing themselves uniformly across the sphere in such a way that they're the furthest away from one another.
$$\rho = Q/V = \frac{Q}{\frac{4\pi}{3} R^3}$$
So to get the encompassed/encircled charge as a function of r:
$$Q_{enc}(r)=\rho V_{enc} = \frac{Q}{\frac{4\pi}{3} R^3} \frac{4\pi}{3} r^3 = Q \frac{r^3}{R^3}$$
We can look at how the integration for inside of the sphere looks now:
$$E4\pi r^2 = \frac{Q\frac{r^3}{R^3} }{\epsilon_0}\\
E = \frac{Q}{4\pi\epsilon_0}\frac{r^3}{R^3r^2} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^3}r
$$
So now to compare the two we have to calculate the E in R/4 and again in 2R. For the first one we use the second formula and we call that E0:
$$E(R/4) = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^3}\frac{R}{4} = \frac{1}{4} \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2} \equiv E_0
$$
The second one at 2R uses the "normal" point charge formula:
$$E(2R) = \frac{Q}{4\pi\epsilon_0r^2} = \frac{Q}{4\pi\epsilon_0(2R)^2}= \frac{1}{4} \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2} $$
which is exactly the same as what we defined E0 to be. So the only correct answer can be c E0 as you say it to be, but I can't get your math.