An isolated spherical capacitor has charge +Q on its inner...

An isolated spherical capacitor has charge $+Q$ on its inner conductor (radius $r_{a} )$ and charge $-Q$ on its outer conductor (radius $r_{b} ) .$ Half of the volume between the two conductors is then filled with a liquid dielectric of constant $K,$ as shown in cross section in Fig. $P 24.78$ . (a) Find the capacitance of the half-filled capacitor. (b) Find the magnitude of $\vec{E}$ in the volume between the two conductors as a function of the distance $r$ from the center of the capacitor. Give answers for both the upper and lower halves of this volume. (c) Find the surface density of free charge on the upper and lower halves of the inner and outer conductors. (d) Find the surface density of bound charge on the inner $\left(r=r_{a}\right)$ and outer $\left(r=r_{b}\right)$ surfaces of the dielectric. (e) What is the surface density of bound charge on the flat surface of the dielectric? Explain.

An isolated spherical capacitor has charge $+Q$ on its inner conductor (radius $r_{a} )$ and charge $-Q$ on its outer conductor (radius $r_{b} ) .$ Half of the volume between the two conductors is
then filled with a liquid dielectric of constant $K,$ as shown in cross section in Fig. $P 24.78$ . (a) Find the capacitance of the half-filled capacitor. (b) Find the magnitude of $\vec{E}$ in the volume between the two conductors as a function of the distance $r$ from the center of the capacitor. Give answers for both the upper and lower halves of this volume. (c) Find the surface density of free charge on the upper and lower halves of the inner and outer conductors. (d) Find the surface density of bound charge on the inner $\left(r=r_{a}\right)$ and outer $\left(r=r_{b}\right)$ surfaces of the dielectric. (e) What is the surface density of bound charge on the flat surface of the dielectric? Explain.

Key Concepts

Electric Field Distribution in Regions with Different Dielectrics

The electric field inside a capacitor may vary spatially when the region between the conductors is filled with materials of different dielectric constants. The field must be computed separately in each region using Gauss’s law, taking into account the distinct permittivities. Continuity conditions at the dielectric interfaces ensure that the displacement field is properly matched, leading to different expressions for the electric field in each region.

Effective Capacitance in Composite Dielectric Systems

When different regions within a capacitor are filled with different materials (or partially filled, as in the half-filled configuration), the overall capacitance is determined by combining the capacitances of individual regions. This often involves treating the regions as connected in series or parallel, depending on their geometry, and requires careful integration of the potential differences across each region.

Polarization and Bound Charges in Dielectrics

Dielectric materials become polarized in an electric field, meaning that microscopic electric dipoles align along the field direction. This polarization gives rise to bound charges, which appear as surface and volume charge densities. The bound surface charge is particularly important at the interfaces (both at the conductor-dielectric boundaries and at discontinuities within the dielectric), affecting the overall electric field distribution and the effective capacitance of the system.

Spherical Capacitor

A spherical capacitor is a system composed of two concentric spherical conductors separated by an insulating medium. Its analysis involves the use of Gauss's law to relate the electric field to the charge enclosed, and the capacitance is determined as the ratio of the charge on one conductor to the potential difference between the conductors. It is a fundamental example in electrostatics, illustrating how geometry influences the energy stored and the electric field distribution.

Dielectric Materials in Capacitors

Dielectric materials are insulating substances placed between the conductors in a capacitor to increase its capacitance. The dielectric constant (relative permittivity) quantifies how much the material reduces the electric field compared to vacuum, thus affecting the overall capacitance. Their presence introduces concepts such as polarization, leading to bound charges, which modify both the electric field and the energy storage characteristics.

Surface Charge Density on Conductors

In capacitors, free charges accumulate on the surfaces of the conductors. The surface charge density is conditioned by the electric field immediately outside the conductors, and in non-uniform dielectric environments, different regions of the conductor surfaces can support different free charge densities. This is crucial for understanding how the stored charge distributes itself in systems with asymmetrical or composite dielectrics.

Transcript

0:00 Hey everybody.

00:01 So for this one, we need to find capsence and then the magnitude and then surface density and then find surface density on charge.

00:10 Okay, so we can find the capsence here as a equals co equals k -epsilon not 2 equals 2 pi epsilon not k, r -a -r -b.

00:30 Divided by r b minus r a okay and get the c u here as c o squared which is going to be 2 pi epsilon not r a b r b are, oops, and this one should be r -b.

01:08 There we go.

01:09 So for this, we got two pi epsilon not, r -a, r -b, r -b minus r -a plus 1, okay? and now for b, okay, we have your e -l is gonna, your el times 4 pi r squared divided by 2 equals ql divided by k -epsilon not.

01:50 E -l equals k -l, i mean ql, 2, k -pire squared, r -squared -epsilon -0.

02:00 And then we have e .u 4 pi r squared 2 equals qu epsilon not.

02:14 Okay.

02:15 And so your eu is going to be qu 2 pi r squared epsilon not.

02:23 Okay.

02:25 And now we have qu equals vcu, which is going to be to v epsilon not squared.

02:34 And your ql is just going to be vkco squared equals k -q -u.

02:47 And if we do q equals those together, so we get k -q -u plus q -u, we get q -u, we get q -u -1 plus q -u, we get q -u -1 plus k and so we have qu equals q over 1 plus k and then we have kl equals kq 1 plus k.

03:18 And so then we have el equals kq equals 2 pi kr squared epsilon knot...

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