leo_africanus
- 4
- 1
- Homework Statement
- We have a capacitor that is made of two concentric conducting spherical shells having radii a and b (b > a). The region between the spheres is filled with a dielectric material which has permittivity $$\epsilon = \frac{k}{r^2}$$
Charge +Q is placed on the inner sphere, while charge -Q is placed on the outer one. We're asked to find the capacitance of the configuration.
- Relevant Equations
- $$ C = \frac{Q}{V} $$
$$ C = \epsilon_r C_{vac} $$
$$ V = - \int E \cdot d\ell$$
$$ D = \epsilon_0 E + P = \epsilon E $$
The first part (which I believe I've done correctly) asks us to find the electric displacement everywhere. For this:
$$\int D \cdot da = Q_{f,enc}$$
For a < r < b: $$D = \frac{Q}{4\pi r^2} \hat{r}$$
Otherwise, D = 0
When finding the capacitance, I'm unsure how to handle the r dependence. I tried two approaches that both yield correct units, but I'm not sure which I believe ( if either ).
1st Attempt: Going the route I've found suggested online, I tried computing the electric field without dielectric present, the corresponding capacitance, and then using $$ C = \epsilon_r C_{vac} = \frac{\epsilon}{\epsilon_0} C_{vac}$$ This yields:
$$ E_{vac} = \frac{Q}{4\pi \epsilon_0 r^2} \hat{r} $$
$$ V_{vac} = \frac{Q}{4\pi \epsilon_0}(\frac{1}{a} - \frac{1}{b}) $$
$$ C_{vac} = 4\pi \epsilon_0 \frac{ab}{b-a}$$
Which gives: $$C = \frac{4\pi k}{r^2} \frac{ab}{b-a} $$
But I'm not sure what a capacitance with lingering radial dependence would mean.
2nd Attempt: Here I just tried computing E in the presence of the dielectric, and then finding V from there, and using C = Q/V. We get:
$$ E = \frac{1}{\epsilon} D = \frac{Q}{4\pi k} \hat{r} $$
$$ V = \frac{Q}{4\pi k} (b-a) $$
Which gives:
$$ C = \frac{4\pi k}{b-a} $$
Is either of these approaches along the right track? They give the same result for a "simpler" permittivity like $$ \epsilon = k \epsilon_0$$ The second approach eliminates the presence of r, but seems inconsistent with approaches I've found online and in text when using dielectric constants that lack r dependence. The first approach matches the second one if I multiply V by the r^2 term before integrating -- which may be the mistake in the first approach. I couldn't find any examples where permittivity has position dependence as here. Any insights are much appreciated! Thanks!
$$\int D \cdot da = Q_{f,enc}$$
For a < r < b: $$D = \frac{Q}{4\pi r^2} \hat{r}$$
Otherwise, D = 0
When finding the capacitance, I'm unsure how to handle the r dependence. I tried two approaches that both yield correct units, but I'm not sure which I believe ( if either ).
1st Attempt: Going the route I've found suggested online, I tried computing the electric field without dielectric present, the corresponding capacitance, and then using $$ C = \epsilon_r C_{vac} = \frac{\epsilon}{\epsilon_0} C_{vac}$$ This yields:
$$ E_{vac} = \frac{Q}{4\pi \epsilon_0 r^2} \hat{r} $$
$$ V_{vac} = \frac{Q}{4\pi \epsilon_0}(\frac{1}{a} - \frac{1}{b}) $$
$$ C_{vac} = 4\pi \epsilon_0 \frac{ab}{b-a}$$
Which gives: $$C = \frac{4\pi k}{r^2} \frac{ab}{b-a} $$
But I'm not sure what a capacitance with lingering radial dependence would mean.
2nd Attempt: Here I just tried computing E in the presence of the dielectric, and then finding V from there, and using C = Q/V. We get:
$$ E = \frac{1}{\epsilon} D = \frac{Q}{4\pi k} \hat{r} $$
$$ V = \frac{Q}{4\pi k} (b-a) $$
Which gives:
$$ C = \frac{4\pi k}{b-a} $$
Is either of these approaches along the right track? They give the same result for a "simpler" permittivity like $$ \epsilon = k \epsilon_0$$ The second approach eliminates the presence of r, but seems inconsistent with approaches I've found online and in text when using dielectric constants that lack r dependence. The first approach matches the second one if I multiply V by the r^2 term before integrating -- which may be the mistake in the first approach. I couldn't find any examples where permittivity has position dependence as here. Any insights are much appreciated! Thanks!