- #1

madafo3435

- 55

- 15

- Homework Statement
- An isolated spherical capacitor has charge + Q in its interior (radius ra) and charge −Q in its outer conductor (radius rb). Then half the volume between the two conductors is filled with a dielectric liquid with constant K. I will attach a picture with the drawing.

Find the capacitance of the half-full capacitor.

- Relevant Equations
- The relevant equations are exposed in the procedure below.

The capacitance must be found, that is:

##C=\frac {Q}{V_a-V_b}##.

Let ##\vec{E_d}## and ##\vec{E_v}## be the electric fields due to the zone with dielectric and the zone without dielectric respectively.

In the case of a spherical capacitor with a vacuum between its plates, it is easy to prove that:

I appreciate the help you can give.

##C=\frac {Q}{V_a-V_b}##.

Let ##\vec{E_d}## and ##\vec{E_v}## be the electric fields due to the zone with dielectric and the zone without dielectric respectively.

In the case of a spherical capacitor with a vacuum between its plates, it is easy to prove that:

##C_0=4\pi \epsilon _0(\frac {r_ar_b}{r_b-r_a})##

Now, here I don't know what to do, but I think the following: If the upper and lower spherical capacitors are at the same potential difference ##V_{ab}##. Then:

##V_a-V_b=\int_{r_a}^{r_b} \vec{E_d} \textbf{.} d\vec{r} = \int_{r_a}^{r_b} \vec{E_v} \textbf{.} d\vec{r} ##

Then the capacitor would be equivalent to two capacitors in parallel, ##C_d## y ##C_v## respectively. Each with half the charge compared to that of the spherical capacitor. Is obtained:

##C_d=\frac {kC_0}{2} = 2\pi K\epsilon _0(\frac {r_ar_b}{r_b-r_a})##

##C_v=\frac {C_0}{2} = 2\pi \epsilon _0(\frac {r_ar_b}{r_b-r_a})##

Therefore:##C_v=\frac {C_0}{2} = 2\pi \epsilon _0(\frac {r_ar_b}{r_b-r_a})##

##C=C_v+C_d=2\pi \epsilon _0(1+K)(\frac {r_ar_b}{r_b-r_a})##

I don't know if this procedure is correct, and if it is, I have trouble proving that the potential difference due to each field mentioned above is the same.

I appreciate the help you can give.