PHIL 1000 Lecture : handout_11_personal_identity.doc

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cyanrabbit590

15 Apr 2013

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It is difficult to see what could account for my identity over time, given that almost all of my properties have changed. Where a1-a9 are actions, q-memory is defined as on p. 362b, and the boxes indicate connectedness. Q-states are defined so as to not beg the question so as to not assume identity. T1 t2: continuity absolutely and connectedness to a factor of 2. T2 t3: continuity absolutely and connectedness to a factor of 3. T1 t3: continuity absolutely and connectedness to a factor of 1. This shows that connectedness, unlike continuity, is a matter of degree. (1) survival of a person over time is a matter of connectedness. (2) connectedness is a matter of degree. ___________________________________________________ (3) survival of a person over time is a matter of degree. It is at best misleading to speak of personal identity over time. we should instead speak of personal survival. _________________________________ (5) a survives as both b and c.

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I am studying for a physics exam next Monday and I need tounderstand this concept a hell of lot better than I do. Hopefullysomeone on here can dumb it down for me.

My textbook says the following:

1. If the probe charge q is positive, the electric fieldvector points in the same direction as the force on the charge; ifnegative, the electric field vector points opposite theforce.

What is this really saying? I don't understand.

2. For the equation, Electric field = K|q|/r^2 , theelectric field is going away from q if q>0, and the electricfield is going toward q if q<0.

I don't understand this either. Is the electric field going awayfrom q because the electric field is negative and q is positive,thus q is attracted to the field? If q is negative then the fieldis going to be going toward q because it's being repelled from thenegative field?

I just want tounderstand the two bolded concepts above and how they relate toeach other, if they in fact do. My job does a horrible job atexplaining it.

Thanks.

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!

I'm pretty lost, i appreciate any help, thankyou

PHIL 1000 Lecture : handout_11_personal_identity.doc

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