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

In summary, the problem asks for the total charge contained in a sphere of radius R centered at the origin, given an electric field of E = kr^2r^. The solution involves using Gauss' Law and integrating over a Gaussian surface with an area element in spherical coordinates. The chosen area vector, directed along the radial direction, leads to the correct answer of 4πε°kR^5. However, another approach could be to use the divergence theorem with a volume element in a ball of radius R. Both methods yield the same result.
  • #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

 
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  • #2
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
You are confused about r versus R, right?
Neolight said:
ds(vector) = r2sinθdθdΦ r∧
How are you defining r there? What surface will you integrate over?
 
  • #4
haruspex said:
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
haruspex said:
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
Neolight said:
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
Neolight said:
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
haruspex said:
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 what's confussing me) .. if we take this form that the answer comes in the format where there is (r)...
 
  • #9
Neolight said:

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

Neolight said:
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 what's 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
haruspex said:
That's all well and good if there are no additional charges in the shell between r and R.
haruspex said:
That's all well and good if there are no additional charges in the shell between r and R.

m
Ray Vickson said:
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
Neolight said:
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
Ray Vickson said:
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
 

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

1. How do I determine the appropriate surface area vector for a flat surface?

The surface area vector for a flat surface is simply a vector perpendicular to the surface. This can be easily determined by using the right-hand rule, where you curl your fingers in the direction of the surface and the vector will point in the direction of your thumb.

2. What is the surface area vector for a curved surface?

The surface area vector for a curved surface is a bit more complicated, as it depends on the geometry of the surface. Generally, it is determined by taking the cross product of two tangent vectors on the surface. However, for simple curved surfaces like a sphere or cylinder, the surface area vector can be determined using geometric formulas.

3. How do I choose the surface area vector for a closed 3D object?

For a closed 3D object, the surface area vector is usually chosen to be outward facing, so that it points away from the inside of the object. This can be determined by using the right-hand rule, where you curl your fingers in the direction of the surface and the vector will point in the direction of your thumb.

4. Do I need to consider the direction of the surface area vector when calculating surface integrals?

Yes, the direction of the surface area vector is important when calculating surface integrals. The direction of the vector determines whether the integral is positive or negative, which affects the final result. Make sure to pay attention to the direction when choosing the surface area vector.

5. Can I choose any surface area vector for a given surface?

No, the surface area vector must be chosen carefully and in accordance with the properties of the surface. Choosing an incorrect surface area vector can result in incorrect calculations and results. It is important to understand the geometry of the surface and choose the appropriate surface area vector for accurate calculations.

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