- #1

vboyn12

- 16

- 2

- Homework Statement:
- In photo below. (help me please)

- Relevant Equations:
- I don't know whether the relevant equation can help because I haven't figure out any idea yet.

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- Thread starter vboyn12
- Start date

- #1

vboyn12

- 16

- 2

- Homework Statement:
- In photo below. (help me please)

- Relevant Equations:
- I don't know whether the relevant equation can help because I haven't figure out any idea yet.

- #2

BvU

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PF rules (please read them) only allow us to help if you post an attempt at solution.

What have you learned so far, that you might use here?

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- #3

vboyn12

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- #4

vboyn12

- 16

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Hi, I try to solve part a and b and i posted below, can you have a look at it?

PF rules (please read them) only allow us to help if you post an attempt at solution.

What have you learned so far, that you might use here?

##\ ##

- #5

BvU

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And then (another one !) $$ \left | \vec E(r_s) \right | = {Q\over 4\pi\varepsilon_0}\ {1\over r_s^2} $$so that $$ \left | \vec E(r_s) \right | = { 12\ \text{V}\ \over \ \displaystyle {1\over r_1} - {1\over r_2}\ } \ {1\over r_s^2} =

12\ \text{V}\ {1\over r_2-r_1}\;{r_1 r_2\over r_s^2 }$$(thus avoiding calculator and rounding errors!) which agrees with your result once we learn to read your handwritten 4 that looks scaringly similar to a 9

I wouldn't quote 6 digits if the ##r## are given in only 1 digit, but that's a matter of taste.

Be sure to use the right ##\varepsilon## if you quote so many digits (I get ##|D| = 1.312 \ 10^{-9} ## !) and

For part c) you need to know that at least the inner sphere is non-conducting !

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- #6

vboyn12

- 16

- 2

And then (another one !) $$ \left | \vec E(r_s) \right | = {Q\over 4\pi\varepsilon_0}\ {1\over r_s^2} $$so that $$ \left | \vec E(r_s) \right | = { 12\ \text{V}\ \over \ \displaystyle {1\over r_1} - {1\over r_2}\ } \ {1\over r_s^2} =

12\ \text{V}\ {1\over r_2-r_1}\;{r_1 r_2\over r_s^2 }$$(thus avoiding calculator and rounding errors!) which agrees with your result once we learn to read your handwritten 4 that looks scaringly similar to a 9

I wouldn't quote 6 digits if the ##r## are given in only 1 digit, but that's a matter of taste.

Be sure to use the right ##\varepsilon## if you quote so many digits (I get ##|D| = 1.312 \ 10^{-9} ## !) anddo use units !

For part c) you need to know that at least the inner sphere is non-conducting !

##\ ##

Sorry about the unit :D. For part (c), is that the value of charge Q found in part (a) equal to the surface charge density multiply by the surface area which is 4pi*r^2? If that is right I get rho(s)=11,789 nC/m^2

And then (another one !) $$ \left | \vec E(r_s) \right | = {Q\over 4\pi\varepsilon_0}\ {1\over r_s^2} $$so that $$ \left | \vec E(r_s) \right | = { 12\ \text{V}\ \over \ \displaystyle {1\over r_1} - {1\over r_2}\ } \ {1\over r_s^2} =

12\ \text{V}\ {1\over r_2-r_1}\;{r_1 r_2\over r_s^2 }$$(thus avoiding calculator and rounding errors!) which agrees with your result once we learn to read your handwritten 4 that looks scaringly similar to a 9

I wouldn't quote 6 digits if the ##r## are given in only 1 digit, but that's a matter of taste.

Be sure to use the right ##\varepsilon## if you quote so many digits (I get ##|D| = 1.312 \ 10^{-9} ## !) anddo use units !

For part c) you need to know that at least the inner sphere is non-conducting !

##\ ##

- #7

BvU

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No problem for me, but it can save you brownie points with graders...Sorry about the unit :D

- #8

vboyn12

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Do you get the same result as mine for part (c)?Up to you: the usual symbol for surface charge is ##\sigma\;##, and ##\rho\ ## is usually volume charge density. Hence my speculation in post #5.

No problem for me, but it can save you brownie points with graders...

- #9

BvU

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What does it say in the actual problem statement ? Is it a conducting sphere or not ?

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- #10

vboyn12

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It is not conductingWhat does it say in the actual problem statement ? Is it a conducting sphere or not ?

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- #11

BvU

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(That is my assumption -- and in that case I don't agree with your answer)

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- #12

vboyn12

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can you give me your solution corresponding to your assumption, please?

(That is my assumption -- and in that case I don't agree with your answer)

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- #13

BvU

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- #14

vboyn12

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Oh, I see :D. Thank you a lot

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