Find electric field intensity/density with given potential difference

[vboyn12]

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.

 

[BvU]

Hi,

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?

$\$

 

[vboyn12]

 

[vboyn12]

Hi,

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?

$\$

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

 

[BvU]

OK, so you use $$ V_1-V_2 = 12\ \text{V}\ = {Q\over 4\pi\varepsilon_0}\left ( {1\over r_1} - {1\over r_2} \right ) $$(which I suppose might make a good entry in the list of relevant equations )

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 do use units !

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

$\$

 

[vboyn12]

OK, so you use $$ V_1-V_2 = 12\ \text{V}\ = {Q\over 4\pi\varepsilon_0}\left ( {1\over r_1} - {1\over r_2} \right ) $$(which I suppose might make a good entry in the list of relevant equations )

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 do use units !

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

$\$

OK, so you use $$ V_1-V_2 = 12\ \text{V}\ = {Q\over 4\pi\varepsilon_0}\left ( {1\over r_1} - {1\over r_2} \right ) $$(which I suppose might make a good entry in the list of relevant equations )

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 do 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

 

[BvU]

Up to you: the usual symbol for surface charge is $\sigma\;$, and $\rho\$ is usually volume charge density. Hence my speculation in post #5.

Sorry about the unit :D

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

 

[vboyn12]

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...

Do you get the same result as mine for part (c)?

 

[BvU]

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

$\$

 

[vboyn12]

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

$\$

It is not conducting

 

[BvU]

And is the charge uniformly distributed over the volume ?
(That is my assumption -- and in that case I don't agree with your answer)

$\$

 

[vboyn12]

And is the charge uniformly distributed over the volume ?
(That is my assumption -- and in that case I don't agree with your answer)

$\$

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

 

[BvU]

You mean calculate the charge density in a uniformly charged unconducting sphere with a given total charge $Q\$ ?$\$

 

[vboyn12]

You mean calculate the charge density in a uniformly charged unconducting sphere with a given total charge $Q\$ ?$\$

Oh, I see :D. Thank you a lot