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 = Q_enclosed/ε_°

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 =Q_enclosed /ε_°

therefore

Q_enclosed = ε_° { ∮ kr⁵ sinθdθdΦ
so after integration we get

Q_enclosed = 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

 

[TSny]

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]

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]

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]

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 what's confussing me) .. if we take this form that the answer comes in the format where there is (r)...

 

[Ray Vickson]

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 = Q_enclosed/ε_°

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 =Q_enclosed /ε_°

therefore

Q_enclosed = ε_° { ∮ kr⁵ sinθdθdΦ
so after integration we get

Q_enclosed = 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 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.

 

[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]

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