Find electric field intensity/density with given potential difference

In summary: Sorry for the mistake. So the surface charge density is equal to the total charge divided by the surface area, which is 4pi*r^2. So the answer should be sigma = 1,170 nC/m^2.
  • #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.
181725852_973626756709481_5184308452962785169_n.jpg
 
Physics news on Phys.org
  • #2
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?

##\ ##
 
  • #3
1.jpg
 
  • #4
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?
 
  • Like
Likes BvU
  • #5
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 !

##\ ##
 
  • #6
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
 
  • #7
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...
 
  • #8
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)?
 
  • #9
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
 

Related to Find electric field intensity/density with given potential difference

1. What is electric field intensity/density?

Electric field intensity/density is a measure of the strength of an electric field at a given point. It is a vector quantity that describes the direction and magnitude of the force that a charge would experience if placed at that point.

2. How is electric field intensity/density related to potential difference?

The electric field intensity/density is directly proportional to the potential difference between two points. This means that a larger potential difference will result in a stronger electric field, and vice versa.

3. What is the formula for calculating electric field intensity/density?

The formula for electric field intensity/density is E = V/d, where E is the electric field intensity/density, V is the potential difference, and d is the distance between the two points.

4. What are the units of electric field intensity/density?

The SI unit for electric field intensity/density is volts per meter (V/m). However, it can also be expressed in other units such as newtons per coulomb (N/C) or volts per centimeter (V/cm).

5. How can I use the formula to find electric field intensity/density with a given potential difference?

To find electric field intensity/density with a given potential difference, simply plug in the values for V and d into the formula E = V/d. Make sure to use consistent units for V and d. The resulting value will be the electric field intensity/density at the given point.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
403
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
278
  • Introductory Physics Homework Help
Replies
11
Views
240
  • Introductory Physics Homework Help
Replies
3
Views
591
  • Introductory Physics Homework Help
Replies
7
Views
465
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
907
Back
Top