Find electric field intensity/density with given potential difference

• vboyn12
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.
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.

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?

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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?

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 !

##\ ##

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 )

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 !

##\ ##
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 )

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

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:
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 ?

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

##\ ##
It is not conducting

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

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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)

##\ ##

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

BvU said:
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

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.

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