zelscore
- 6
- 0
- 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.
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.