- #1
YehiaMedhat
- 21
- 3
- Homework Statement
- A hemispherical shell of radius ##4.0 cm## is located at the region ##z > 0.0## centered at the origin. It carries a surface charge distribution of density ##\rho = 2.0\times 10^{-6} C/m^2## in the region ##y > 0.0## and a surface charge distribution of density ##\rho = -2.0\times 10^{-6} C/m^2## in the region ##y < 0.0##. Apply Coulomb's law to find the electric field intensity ##E## at the origin.
- Relevant Equations
- ## \vec{E} = \frac{Q}{4\pi \epsilon r^2} \vec{a} ##
The question is:
A hemispherical shell of radius ##4.0 cm## is located at the region ##z > 0.0## centered at the origin. It carries a surface charge distribution of density ##\rho = 2.0\times 10^{-6} C/m^2## in the region ##y > 0.0## and a surface charge distribution of density ##\rho = -2.0\times 10^{-6} C/m^2## in the region ##y < 0.0##. Apply Coulomb's law to find the electric field intensity ##E## at the origin.
For me I considered slicing each half of the hemisphere into rings to get an intuition about how will the electric field will look like, so as in the picture, each vertical component is gonna be canceled with the other one from the opposing charge area, so we're left only with the horizontal components multiplied by 2.
Integrating with phi from 0 to pi over 2 and theta from 0 to 2pi
##
\begin{aligned}
& 2 \int_0^{2 \pi} \int_0^{\frac{\pi}{2}} \frac{(\rho d \phi)(f d \theta) \rho_s(\cos \phi)}{4 \pi \varepsilon_0 \rho^2} \\
& =\frac{2 \rho_s}{4 \pi \varepsilon_0} \int_0^{2 \pi} d \theta \int_0^{\frac{\pi}{2}} \cos \phi d \phi \\
& =\frac{(2 \pi) 2 \rho}{4 \pi \varepsilon_0}[\sin \phi]_0^{\frac{\pi}{2}}=\frac{\rho_s}{\varepsilon_0}
\end{aligned}
##
Is that correct?
A hemispherical shell of radius ##4.0 cm## is located at the region ##z > 0.0## centered at the origin. It carries a surface charge distribution of density ##\rho = 2.0\times 10^{-6} C/m^2## in the region ##y > 0.0## and a surface charge distribution of density ##\rho = -2.0\times 10^{-6} C/m^2## in the region ##y < 0.0##. Apply Coulomb's law to find the electric field intensity ##E## at the origin.
For me I considered slicing each half of the hemisphere into rings to get an intuition about how will the electric field will look like, so as in the picture, each vertical component is gonna be canceled with the other one from the opposing charge area, so we're left only with the horizontal components multiplied by 2.
Integrating with phi from 0 to pi over 2 and theta from 0 to 2pi
##
\begin{aligned}
& 2 \int_0^{2 \pi} \int_0^{\frac{\pi}{2}} \frac{(\rho d \phi)(f d \theta) \rho_s(\cos \phi)}{4 \pi \varepsilon_0 \rho^2} \\
& =\frac{2 \rho_s}{4 \pi \varepsilon_0} \int_0^{2 \pi} d \theta \int_0^{\frac{\pi}{2}} \cos \phi d \phi \\
& =\frac{(2 \pi) 2 \rho}{4 \pi \varepsilon_0}[\sin \phi]_0^{\frac{\pi}{2}}=\frac{\rho_s}{\varepsilon_0}
\end{aligned}
##
Is that correct?