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184_notes:examples:week5_flux_two_radii [2017/09/18 13:20] – [Example: Flux through Two Spherical Shells] tallpaul | 184_notes:examples:week5_flux_two_radii [2018/07/24 15:02] – curdemma | ||
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=====Example: | =====Example: | ||
Suppose you have a point charge with value $1 \mu\text{C}$. What are the fluxes through two spherical shells centered at the point charge, one with radius $3 \text{ cm}$ and the other with radius $6 \text{ cm}$? | Suppose you have a point charge with value $1 \mu\text{C}$. What are the fluxes through two spherical shells centered at the point charge, one with radius $3 \text{ cm}$ and the other with radius $6 \text{ cm}$? | ||
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{{ 184_notes: | {{ 184_notes: | ||
====Solution==== | ====Solution==== | ||
- | Before we dive into calculations, | + | Before we dive into calculations, |
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+ | {{ 184_notes: | ||
- | Since $\vec{E}$ is constant with respect | + | Since the shell is a fixed distance from the point charge, the electric field has constant magnitude on the shell. |
- | $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} = \left|\vec{E}\right|\int\text{d}\left|\vec{A}\right|$$ | + | $$\Phi_e=\int\vec{E}\bullet \text{d}\vec{A} = E\int\text{d}A$$ |
- | We can rewrite | + | Note that $E$ is a scalar value representing |
- | $$\left|\vec{E}\right| | + | $$E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}\left|\hat{r}\right| = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$$ |
- | We can plug in values for $q$ and $r$ for each spherical shell, using what we listed in the facts. For the smaller shell, we have $\left|\vec{E}\right|=1.0\cdot 10^7 \text{ N/C}$. For the larger shell, we have $\left|\vec{E}\right|=2.5\cdot 10^6 \text{ N/C}$. | + | We can plug in values for $q$ and $r$ for each spherical shell, using what we listed in the facts. For the smaller shell, we find $E=1.0\cdot 10^7 \text{ N/C}$. For the larger shell, we find $E=2.5\cdot 10^6 \text{ N/C}$. |
To figure out the area integral, notice that the magnitude of the area-vector is just the area. This means that our integrand is the area occupied by $\text{d}A$. Since we are integrating this little piece over the entire shell, we end up with the area of the shell' | To figure out the area integral, notice that the magnitude of the area-vector is just the area. This means that our integrand is the area occupied by $\text{d}A$. Since we are integrating this little piece over the entire shell, we end up with the area of the shell' | ||
- | $$\int\text{d}\left|\vec{A}\right|=\int\text{d}A=A=4\pi r^2$$ | + | $$\int\text{d}A=A=4\pi r^2$$ |
The last expression, $4\pi r^2$, is just the surface area of a sphere. We can plug in values for $r$ for each spherical shell, using what we listed in the facts. For the smaller shell, we have $A=1.13\cdot 10^{-2} \text{ m}^2$. For the larger shell, we have $A=4.52\cdot 10^{-2} \text{ m}^2$. | The last expression, $4\pi r^2$, is just the surface area of a sphere. We can plug in values for $r$ for each spherical shell, using what we listed in the facts. For the smaller shell, we have $A=1.13\cdot 10^{-2} \text{ m}^2$. For the larger shell, we have $A=4.52\cdot 10^{-2} \text{ m}^2$. | ||
- | Now, we bring it together to find electric flux, which after all our simplifications can be written as $\Phi_e=\left|\vec{E}\right|A$. For our two shells: | + | Now, we bring it together to find electric flux, which after all our simplifications can be written as $\Phi_e=EA$. For our two shells: |
\begin{align*} | \begin{align*} | ||
\Phi_{\text{small}} &= 1.0\cdot 10^7 \text{ N/C } \cdot 1.13\cdot 10^{-2} \text{ m}^2 = 1.13 \cdot 10^5 \text{ Nm}^2\text{/ | \Phi_{\text{small}} &= 1.0\cdot 10^7 \text{ N/C } \cdot 1.13\cdot 10^{-2} \text{ m}^2 = 1.13 \cdot 10^5 \text{ Nm}^2\text{/ |