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

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

is there a mistake in the selection of area vector?

## The Attempt at a Solution

TSny
Homework Helper
Gold Member
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?

haruspex
Homework Helper
Gold Member
2020 Award
You are confused about r versus R, right?
ds(vector) = r2sinθdθdΦ r∧
How are you defining r there? What surface will you integrate over?

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

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..

haruspex
Homework Helper
Gold Member
2020 Award
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.

haruspex
Homework Helper
Gold Member
2020 Award
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.

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)...

Ray Vickson
Homework Helper
Dearly Missed

## 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

is there a mistake in the selection of area vector?

## 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.

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?

Ray Vickson
Homework Helper
Dearly Missed
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.

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.
sorry about that , I won't make that mistake next time