Find electric field intensity/density with given potential difference

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SUMMARY

The forum discussion focuses on calculating electric field intensity and charge density from a given potential difference of 12 V using the equations for electric potential and electric field. The relevant equations discussed include $$ V_1-V_2 = 12\ \text{V} = {Q\over 4\pi\varepsilon_0}\left ( {1\over r_1} - {1\over r_2} \right ) $$ and $$ \left | \vec E(r_s) \right | = {Q\over 4\pi\varepsilon_0}\ {1\over r_s^2} $$. The discussion emphasizes the importance of using correct units and symbols, particularly distinguishing between surface charge density (σ) and volume charge density (ρ). Participants also clarify that the inner sphere is non-conducting, which is crucial for accurate calculations.

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  • Understanding of electric potential and electric field concepts
  • Familiarity with the equations of electrostatics, particularly Gauss's Law
  • Knowledge of charge density definitions and differences between surface and volume charge densities
  • Ability to manipulate mathematical equations involving constants like ε₀ (permittivity of free space)
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  • Learn how to calculate electric field intensity from potential difference in various geometries
  • Research the implications of charge distribution in non-conducting materials
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vboyn12
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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.
181725852_973626756709481_5184308452962785169_n.jpg
 
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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?

##\ ##
 
1.jpg
 
BvU said:
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?
 
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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 :rolleyes: )

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 :nb)

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 ! :mad:

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

##\ ##
 
BvU said:
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 :rolleyes: )

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 :nb)

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 ! :mad:

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

##\ ##
BvU said:
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 :rolleyes: )

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 :nb)

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 ! :mad:

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


vboyn12 said:
Sorry about the unit :D
No problem for me, but it can save you brownie points with graders...
 
BvU said:
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)?
 
What does it say in the actual problem statement ? Is it a conducting sphere or not ?

##\ ##
 
  • #10
BvU said:
What does it say in the actual problem statement ? Is it a conducting sphere or not ?

##\ ##
It is not conducting
 
  • #11
And is the charge uniformly distributed over the volume ?
(That is my assumption -- and in that case I don't agree with your answer)

##\ ##
 
  • #12
BvU said:
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?
 
  • #13
You mean calculate the charge density in a uniformly charged unconducting sphere with a given total charge ##Q\ ## :wink: ?##\ ##
 
  • #14
BvU said:
You mean calculate the charge density in a uniformly charged unconducting sphere with a given total charge ##Q\ ## :wink: ?##\ ##
Oh, I see :D. Thank you a lot
 

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