How do i choose the Surface area vector in different cases?

  • Thread starter Neolight
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  • #1
Neolight

Homework Statement


Suppose we have a Electic field, E (vector) = kr2r(vector). Find the total charge contained in the sphere of radius R centered in the origin.
solution-
here E can also be written as kr3r, where r is the unit vector
this is the given question , so obiously the best way to solve this is to use Gauss Law but while using it i have a confusion on which surface area vector to choose..

since gauss law is
∮E.ds = Qenclosed°

this is where my confussion starts ,ds(surface area vector in spherical polar ) = r2sin(θ )drdθdΦ
but here the area element and E are in dot product so i have to use vector form of ds
since this is a sphere the area vector will be directed along the radial line so
ds(vector) = r2sinθdθdΦ r

so after putting this in the equation and doing dot product i get
∮(kr3 )( r2sinθdθdΦ =Qenclosed /ε°

therefore

Qenclosed = ε° { ∮ kr5 sinθdθdΦ
so after integration we get

Qenclosed = 4πε°kr5

but somehow in the answer given in the book it is
4πε°kR5

what am i doing wrong here ? please help
is there a mistake in the selection of area vector?

Homework Equations




The Attempt at a Solution

 

Answers and Replies

  • #2
TSny
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If you want to find the total charge inside a sphere of radius R, what should you choose for the radius r of your Gaussian surface?
 
  • #3
haruspex
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You are confused about r versus R, right?
ds(vector) = r2sinθdθdΦ r∧
How are you defining r there? What surface will you integrate over?
 
  • #4
Neolight
You are confused about r versus R, right?

How are you defining r there? What surface will you integrate over?
Yes.. I know there is something wrong with what I'm doing .. but just can't get what it is
 
  • #5
Neolight
You are confused about r versus R, right?

How are you defining r there? What surface will you integrate over?
well r as the radius of a gaussian sphere where r>R..
 
  • #6
haruspex
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Yes.. I know there is something wrong with what I'm doing .. but just can't get what it is
You did not answer my questions.
 
  • #7
haruspex
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well r as the radius of a gaussian sphere where r>R..
That's all well and good if there are no additional charges in the shell between r and R.
 
  • #8
Neolight
That's all well and good if there are no additional charges in the shell between r and R.
the thing i'm confused is that area vector, since the problem is for a sphere we need to take spherical coordinate since it will be much more easier.
and the area vector i used here is directed towards the radial direction as dr=0 ( here is whats confussing me) .. if we take this form that the answer comes in the format where there is (r)...
 
  • #9
Ray Vickson
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Homework Statement


Suppose we have a Electic field, E (vector) = kr2r(vector). Find the total charge contained in the sphere of radius R centered in the origin.
solution-
here E can also be written as kr3r, where r is the unit vector
this is the given question , so obiously the best way to solve this is to use Gauss Law but while using it i have a confusion on which surface area vector to choose..

since gauss law is
∮E.ds = Qenclosed°

this is where my confussion starts ,ds(surface area vector in spherical polar ) = r2sin(θ )drdθdΦ
but here the area element and E are in dot product so i have to use vector form of ds
since this is a sphere the area vector will be directed along the radial line so
ds(vector) = r2sinθdθdΦ r

so after putting this in the equation and doing dot product i get
∮(kr3 )( r2sinθdθdΦ =Qenclosed /ε°

therefore

Qenclosed = ε° { ∮ kr5 sinθdθdΦ
so after integration we get

Qenclosed = 4πε°kr5

but somehow in the answer given in the book it is
4πε°kR5

what am i doing wrong here ? please help
is there a mistake in the selection of area vector?

Homework Equations




The Attempt at a Solution

the thing i'm confused is that area vector, since the problem is for a sphere we need to take spherical coordinate since it will be much more easier.
and the area vector i used here is directed towards the radial direction as dr=0 ( here is whats confussing me) .. if we take this form that the answer comes in the format where there is (r)...

You could also use
$$\oint_{S_R} \mathbf{E} \cdot d\mathbf{S} = \int_{B_R} \text{div}{\mathbf{E}} \, dV ,$$
where ##B_R## is the ball of radius ##R##, ##S_R## is its surface and ##dV## is the volume element.
 
  • #10
Neolight
That's all well and good if there are no additional charges in the shell between r and R.
That's all well and good if there are no additional charges in the shell between r and R.

m
You could also use
$$\oint_{S_R} \mathbf{E} \cdot d\mathbf{S} = \int_{B_R} \text{div}{\mathbf{E}} \, dV ,$$
where ##B_R## is the ball of radius ##R##, ##S_R## is its surface and ##dV## is the volume element.
ok i'll try that , but what do u think about the surface element i took was it correct method?
 
  • #11
Ray Vickson
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m

ok i'll try that , but what do u think about the surface element i took was it correct method?

Sure, it is OK.

I assume your "i" means "I" (not ##\sqrt{-1}##) and that your "u" means "you". Please avoid textspeak in this forum.
 
  • #12
Neolight
Sure, it is OK.

I assume your "i" means "I" (not ##\sqrt{-1}##) and that your "u" means "you". Please avoid textspeak in this forum.
:smile::smile: sorry about that , I won't make that mistake next time
 

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