Find the electric field on the surface of a sphere using Coulomb's law

[zelscore]

Homework Statement

A charged hollow metal sphere of negligible thickness, with outer radius 0.04m has its center positioned at the origin (0,0,0) of a Cartesian coordinate system. The sphere is loaded with a charge of 1 nC. It is surrounded by free space.

Calculate the electric field (magnitude and direction) at the point (0, 0, 0.04) precisely outside the surface of the sphere.

Relevant Equations

Coulombs law: E = $\frac {1} {4πε}$ $\int$ $\frac {\rho R_{SO}} {R^3_{SO}} \, dv$
$R_{SO}$ is a vector pointing from the source point to the observation point. $R_{SO} = R_O - R_S$

$\epsilon = 8.85 * 10^-12$

Note that the solution is 5625 V/m in z direction which is found easier using Gauss' law, but I want to find the same result using Coulombs law for confirmation.

Lets give the radius 0.04 the variable a = 0.04m.
$\rho$ is the charge distribution distributed evenly on the surface of the sphere, which is equal to Q/Area = $\frac {1*10^{-9}} {4 \pi a^2}$ = $4.97*10^{-8}$

We find $R_{SO} = R_O - R_S$ = $(0, 0, a) - (asin \theta cos\phi, asin \theta sin \phi, acos \theta) = (-asin \theta cos \phi, -asin \theta sin \phi, a - acos \theta)$

Note that $R_S$ is just the vector pointing from origo to a point on the sphere, so it is given by the spherical coordinates above.

Now $R^3_{SO}$ = magnitude of $R_{SO}$ cubed $= \sqrt{a^2sin^2 \theta cos^2 \phi + a^2 sin^2 \theta sin^2 \phi + a^2 + a^2 cos^2 \theta}^3$ = $\sqrt{a^2 + a^2}^3 = 0.00181$ by use of the trigonometric identity $sin^2 x + cos^2 x = 1$

The Jacobian for spherical coordinates is $a^2 sin \theta$

Now I put everything into Coulombs law and E = $\frac {1} {4πε}$ $\int$ $\frac {\rho R_{SO}} {R^3_{SO}}$ $\, dv$ = $\frac {\rho} {4 \pi \epsilon}$ $\int_0^{2\pi}$ $\int_0^\pi$ $\frac {(-asin \theta cos \phi, -asin \theta sin \phi, a - acos \theta)} {0.000181}$ $0.04^2 sin \theta \, d \theta d \phi$ = $$\frac {\rho 0.04^2} {4 \pi \epsilon 0.000181} \int_0^{2\pi} \int_0^\pi (-asin \theta cos \phi, -asin \theta sin \phi, a - acos \theta) sin \theta \, d \theta d \phi =$$

$$ 3950 \int_0^{2\pi} \int_0^\pi (-asin^2 \theta cos \phi, -asin^2 \theta sin \phi, a sin \theta - acos \theta sin \theta) \, d \theta d \phi $$

However I get stuck at this step, can someone help me solve this or help anyhow i'd appreciate.

 

[PeroK]

Homework Statement:: A charged hollow metal sphere of negligible thickness, with outer radius 0.04m has its center positioned at the origin (0,0,0) of a Cartesian coordinate system. The sphere is loaded with a charge of 1 nC. It is surrounded by free space.

Calculate the electric field (magnitude and direction) at the point (0, 0, 0.04) precisely outside the surface of the sphere.
Homework Equations:: Coulombs law: E = $\frac {1} {4πε}$ $\int$ $\frac {\rho R_{SO}} {R^3_{SO}} \, dv$
$R_{SO}$ is a vector pointing from the source point to the observation point. $R_{SO} = R_O - R_S$

$\epsilon = 8.85 * 10^-12$

Note that the solution is 5625 V/m in z direction which is found easier using Gauss' law, but I want to find the same result using Coulombs law for confirmation.

Lets give the radius 0.04 the variable a = 0.04m.
$\rho$ is the charge distribution distributed evenly on the surface of the sphere, which is equal to Q/Area = $\frac {1*10^{-9}} {4 \pi a^2}$ = $4.97*10^{-8}$

We find $R_{SO} = R_O - R_S$ = $(0, 0, a) - (asin \theta cos\phi, asin \theta sin \phi, acos \theta) = (-asin \theta cos \phi, -asin \theta sin \phi, a - acos \theta)$

Note that $R_S$ is just the vector pointing from origo to a point on the sphere, so it is given by the spherical coordinates above.

Now $R^3_{SO}$ = magnitude of $R_{SO}$ cubed $= \sqrt{a^2sin^2 \theta cos^2 \phi + a^2 sin^2 \theta sin^2 \phi + a^2 + a^2 cos^2 \theta}^3$ = $\sqrt{a^2 + a^2}^3 = 0.00181$ by use of the trigonometric identity $sin^2 x + cos^2 x = 1$

The Jacobian for spherical coordinates is $a^2 sin \theta$

Now I put everything into Coulombs law and E = $\frac {1} {4πε}$ $\int$ $\frac {\rho R_{SO}} {R^3_{SO}}$ $\, dv$ = $\frac {\rho} {4 \pi \epsilon}$ $\int_0^{2\pi}$ $\int_0^\pi$ $\frac {(-asin \theta cos \phi, -asin \theta sin \phi, a - acos \theta)} {0.000181}$ $0.04^2 sin \theta \, d \theta d \phi$ = $$\frac {\rho 0.04^2} {4 \pi \epsilon 0.000181} \int_0^{2\pi} \int_0^\pi (-asin \theta cos \phi, -asin \theta sin \phi, a - acos \theta) sin \theta \, d \theta d \phi =$$

$$ 3950 \int_0^{2\pi} \int_0^\pi (-asin^2 \theta cos \phi, -asin^2 \theta sin \phi, a sin \theta - acos \theta sin \theta) \, d \theta d \phi $$

However I get stuck at this step, can someone help me solve this or help anyhow i'd appreciate.

What precisely is stopping you from integrating there?

 

[ehild]

Now $R^3_{SO}$ = magnitude of $R_{SO}$ cubed $= \sqrt{a^2sin^2 \theta cos^2 \phi + a^2 sin^2 \theta sin^2 \phi + a^2 + a^2 cos^2 \theta}^3$

$a^2 + a^2 \cos^2 \theta$ is wrong, it should be the square of $a-a\cos(\theta)$.

 

[zelscore]

$a^2 + a^2 \cos^2 \theta$ is wrong, it should be the square of $a-a\cos(\theta)$.

Okay I corrected this and get $R^3_{SO}$ to equal $\sqrt {a^2 (2-2cos \theta)}^3$ instead of 0.000181. However here is a picture showing an integral calculator online failing to solve this integral. Any ideas? This problem is really confusing me because considering I am working with a sphere, there should be some geometric symmetry which makes for "good looking" numbers as a result, not an unsolvable integral...

 

[zelscore]

What precisely is stopping you from integrating there?

Okay I actually made an error like the comment by another user pinpointed, but I corrected this and get $R^3_{SO}$ to equal $\sqrt {a^2 (2-2cos \theta)}^3$ instead of 0.000181. However here is a picture showing an integral calculator online failing to solve this integral. Any ideas? This problem is really confusing me because considering I am working with a sphere, there should be some geometric symmetry which makes for "good looking" numbers as a result, not an unsolvable integral...

 

[PeroK]

Any ideas? This problem is really confusing me because considering I am working with a sphere, there should be some geometric symmetry which makes for "good looking" numbers as a result, not an unsolvable integral...

That integral is not hard to do. It's only sines and cosines.

 

[ehild]

Okay I corrected this and get $R^3_{SO}$ to equal $\sqrt {a^2 (2-2cos \theta)}^3$ instead of 0.000181. However here is a picture showing an integral calculator online failing to solve this integral.

Use the identity 1-cos´(theta)=0.5(sin^2(theta/2). Write the integral separately for the x,y,z components of the electric field.