PHYSICS 3B Chapter Notes - Chapter 21: Gaussian Surface, Electric Flux, Power Rangers Zeo

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pinkhornet528

29 Mar 2020

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Oh , 22 : gauss "s law v hux m . Electric tux through an area is proportional to the total number of field lines crossing the area. Electric flux through a closed surface : The net number of field lines through the summer is proportional and also the electric flux. This can be used to find the elective field in a situation where the charge distribution has a high degree of spatial symmetry (cid:8869) ex . charge distributed on a sphere or along a long wire. For a point charge , f e -dat = * gauss"s law is another way of stating coulomb "s law! Solving for e gives us coulomb "s law : e - afar, Result is valid for any shape of a closed surface integral. , but symmetry makes it easierto do the. " ifa gaussian surface encloses several point charges , the superposition principle shows that.

Related Questions

A particle of mass m is constrained to move on the inside surfaceof a cone of half-angle a, under the influence of constantgravitational force. In the cylindrical coordinates (r, phi, z) theconstraint that the particle is moving on the surface of the coneis given by z=r cot a . [do not use lagrangian mothod)

a) Find the net force acting on the particle and derive theequation of motion for r and phi using Newton's 2nd law. Show thatthe angular momentum L is a constant of motion. (use a incylindrical coordinateshttp://mathworld.wolfram.com/CylindricalCoordinates.html
see eq 72)
b) Find the total energy in terms of r, r dot and L

My attempt:

a) I found all the forces acting on the particle and set them equalto the forces I get from using the acceleration in cylindricalcoordinates

so, F(in phi direction) = 0 = m (2(r dot)(phi dot) + r (phidoubledot))
F(r) = -N cos a = m( (r double dot - r* (phi dot)^2) [N = normalforce]
F(z) = -mg + N sin a = m (z double dot)

How should I proceed next ? I have to find the equations of motionfor r and phi.
And how should I show that L is a constant of motion?

b) I have no idea how to do this part!

Part A:
A capacitor is completely charged with 650 nC by a voltagesourcethat has 275 V.
What is its capacitance?

C=Q/V = (550e-9 C)/375V = 1/47e-9 F (1.47 nF)ConfirmedRight

Part B:
Now the plates of the charged capacitor are pushed together withthevoltage source already disconnected:

- The voltage drop between the plates decreases.
- The energy stored in the capacitor remains the same.
- The capacitance increases.
- The voltage drop between the plates increases.
- None of the above.

Part C:
The initial air gap of the capacitor above was 7 mm. What isthestored energy if the air gap is now 3 mm?

1) We know Capacitance when the air gap is 7 mm (d1=7e-3 m), butnotthe area, A. So use:

C= (Îµ0A)/(d1)
or
A=(Cd)/(Îµ0) = ((1.47E-9 F)*(7e-3 m))/(8.85e-12) = 1.163m^2

2) Using the area, A, that you find, find the capacitance, C'withan air gap of 3 mm. (d2=3e-3 m)

C'= (Îµ0A)/(d2) = ((8.85e-12)*(1.163 m^2))/(3e-3) = 3.431e-9F

3) The potential energy with this air gap of 3 mm is:

U = 0.5C'V^2 = 0.5(3.431e-9 F)(375^2) = 2.41e-4 JNotRight

Alright, as you can see, I worked out the last part completelyandI'm pretty sure my work is right. However, the answer is showingupas wrong. I'm not sure what I did wrong here. Also, I'm notsurewhat to do for Part B and I only have one try. Wouldn't theenergyremain the same (or nothing happen at all)?

Need some help!

A block of mass m = 2 kg has a speed V and is behind a block ofmassM = 44 kg that has a speed of 0.5 m/s. The surface isfrictionless.The blocks collide and couple. After the collision,the blocks havea common speed of 0.9 m/s. In Fig. 8.2, the impulseon the 2-kgblock due to the collision is closest to:

0.80 {N}\cdot{s}
40 {N}\cdot{s}
22 {N}\cdot{s}
31 {N}\cdot{s}
18 {N}\cdot{s}

Now i have worked the math and got that the speed V was 9.7 butwhenit comes time to calculate the impulse which is p2-p1 i don'tknowif it is the momentum of the two objects together or is itonly ofthe 2kg object.

i calculated 41.4-19.4 and got 22N but that turns out to bethewrong answer so then i did 1.8-19.4 and got -17.6 but that isnoteven one of the choices i have been given. i am pretty sure iamjust doing some stupid mistake can somebody help me.

PHYSICS 3B Chapter Notes - Chapter 21: Gaussian Surface, Electric Flux, Power Rangers Zeo

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