Cylindrical Capacitors and Electric Fields

2012 Test 2 Q3

Current density J and electric field E are related by conductivity \,sigma: J = \,sigma E.
E field in the z-direction: E = -\frac{V}{h} \hat{z} (V/m).
J field in the z-direction: J = \frac{I}{S} \hat{z} (A/m^2), where S is the cross-sectional area.
Cross-sectional area S is given by: S = \pi (b^2 - a^2).
Therefore, J = \frac{I_0}{\pi (b^2 - a^2)} \hat{z}.

Potential difference V is calculated by integrating the electric field along a path:
V = -\int E \cdot dl.
E field: E = -\frac{I_0}{2 \pi \,sigma r h} \hat{r}.
The potential difference between points a and b:
V = - \inta^b E \cdot dl = - \inta^b \frac{I0}{2 \pi \,sigma r h} dr \hat{r} \cdot \hat{r} = - \frac{I0}{2 \pi \,sigma h} \int_a^b \frac{1}{r} dr.
Evaluating the integral:
V = - \frac{I0}{2 \pi \,sigma h} [\ln(r)]a^b = - \frac{I0}{2 \pi \,sigma h} (\ln(b) - \ln(a)) = \frac{I0}{2 \pi \,sigma h} \ln(\frac{a}{b}).

Resistance is defined as: R = \frac{V}{I}.
Given V = V_0 and parameters: h = 99.75 mm, (\frac{b}{a}) = 190, we need to find R.
R = \frac{V}{I} = \frac{h}{\pi \,sigma (b^2 - a^2)}.

Given V(r, \phi, z) = V(z), the potential only varies with z.
Laplace's equation in this case simplifies to: \nabla^2 V = \frac{\partial^2 V}{\partial z^2} = 0.

Boundary conditions are:
- V(z = 0) = V0. This gives us K2 = V_0.
- V(z = h) = Vh. This gives us K1 + K2 = Vh.
The general solution for V(z) is V(z) = K1 z + K2. Using the boundary conditions:
V(z) = \frac{Vh - V0}{h}z + V_0.

Electric field is the negative gradient of the potential: E = - \nabla V = - \frac{\partial V}{\partial z} \hat{z}.
E = - \frac{Vh - V0}{h} \hat{z} = \frac{V0 - Vh}{h} \hat{z} (V/m).

Current is the integral of current density over the cross-sectional area:
I = \int J \cdot dS = \int0^{2\pi} \inta^b \frac{\,sigma (V0 - Vh)}{h} r dr d\phi \hat{z} \cdot \hat{z}.
I = \frac{\,sigma (V0 - Vh)}{h} \int0^{2\pi} d\phi \inta^b r dr = \frac{\,sigma (V0 - Vh)}{h} (2 \pi) [\frac{1}{2} r^2]_a^b.
I = \frac{\,sigma \pi (V0 - Vh) (b^2 - a^2)}{h}.

Problem setup: Infinitely long hollow cylindrical volume with uniform volume charge density \,rhov0 between r = a and r = b, where a < b and \,rhov0 is a positive constant.
- Find the electric field intensity E(r) for a < r < b and r > b.
- Determine the potential difference V_{ab} between cylindrical surfaces at r = a and r = b.
- Find the required uniform line charge density \,rho_l on the z-axis to ensure E(r) = 0 for r > b.

Enclosed charge: Q{incl} = \intV \rhov dV = \int0^h \int0^{2\pi} \inta^r \rhov0 r dr d\phi dz = \rhov0 h \pi (r^2 - a^2).
Applying Gauss's law: E(r) 2 \pi r h = \frac{\rhov0 \pi h (r^2 - a^2)}{\epsilon_0}.
E(r) = \frac{\rhov0 (r^2 - a^2)}{2 \epsilon_0 r} \hat{r} V/m.

Enclosed charge: Q{incl} = \intV \rhov dV = \int0^h \int0^{2\pi} \inta^b \rhov0 r dr d\phi dz = \rhov0 h \pi (b^2 - a^2).
Applying Gauss's law:
E(r) 2 \pi r h = \frac{\rhov0 \pi h (b^2 - a^2)}{\epsilon_0}.
E(r) = \frac{\rhov0 (b^2 - a^2)}{2 \epsilon_0 r} \hat{r} V/m.

V{ab} = - \inta^b E \cdot dl = - \inta^b \frac{\rhov0 (r^2 - a^2)}{2 \epsilon0 r} dr \hat{r} \cdot \hat{r} = - \frac{\rhov0}{2 \epsilon0} \inta^b (r - \frac{a^2}{r}) dr.
V{ab} = - \frac{\rhov0}{2 \epsilon0} [\frac{r^2}{2} - a^2 \ln(r)]a^b = \frac{\rhov0}{2 \epsilon0} [\frac{a^2}{2} - a^2 \ln(a) - \frac{b^2}{2} + a^2 \ln(b)] = \frac{\rhov0}{2 \epsilon_0} (\frac{a^2 - b^2}{2} + a^2 \ln(\frac{b}{a})).

To ensure E(r) = 0 for r > b, the total enclosed charge must be zero.
Q{incl} = Qv + Ql = 0, where Qv is the volume charge and Q_l is the line charge.
Qv = \rhov_0 \pi (b^2 - a^2) h.
Ql = \rhol h.
Therefore, \,rhol h + \rhov_0 \pi (b^2 - a^2) h = 0.
\,rhol = - \rhov_0 \pi (b^2 - a^2) C/m.

Problem setup: Cylindrical capacitor of length l with radii r = a and r = b at z = 0 and z = l, respectively. The radius varies linearly from z = 0 to z = l.
The space between the terminals is filled with a lossy material (permittivity \,epsilon and conductivity \,sigma).
- Find the current density J(z).
- Find the potential difference between the terminals.
- Find the DC resistance.

The current I flows in the z-direction.
J = \,sigma E.
The radius r varies linearly with z: r(z) = a + \frac{b - a}{l} z.
Area: S = \pi r^2 = \pi (a + \frac{b - a}{l} z)^2.
J(z) = \frac{I}{S} \hat{z} = \frac{I}{\pi (a + \frac{b - a}{l} z)^2} \hat{z} A/m^2.

E = \frac{J}{\,sigma} = \frac{I}{\,sigma \pi (a + \frac{b - a}{l} z)^2} \hat{z}.
V = - \int E \cdot dl = - \int_0^l \frac{I}{\,sigma \pi (a + \frac{b - a}{l} z)^2} dz \hat{z} \cdot \hat{z}.
Let c = \frac{b - a}{l}. Then V = - \frac{I}{\,sigma \pi} \int_0^l \frac{1}{(a + cz)^2} dz.
V = - \frac{I}{\,sigma \pi} [-\frac{1}{c(a + cz)}]_0^l = \frac{I}{\,sigma \pi c} [\frac{1}{a + cl} - \frac{1}{a}] = \frac{I}{\,sigma \pi (\frac{b - a}{l})} [\frac{1}{b} - \frac{1}{a}].
V = \frac{Il}{\,sigma \pi (b - a)} [\frac{a - b}{ab}] = \frac{Il (a - b)}{\,sigma \pi (b - a) ab} = -\frac{Il}{\,sigma \pi ab}.

R = \frac{V}{I} = \frac{- \frac{Il}{\,sigma \pi ab}}{I} = \frac{l}{\,sigma \pi ab}.

Problem setup: Cylindrical capacitor of length 2l with radius=b at z = -l and z = l, respectively. The radius varies linearly from r = a at z = 0 to r = b at z = \pm l.
The space between the two conducting terminals is homogeneously filled with a lossy material (permittivity \,epsilon and conductivity \,sigma).
- Find the potential difference between the terminals,
- and the DC resistance due to the lossy material between the two conducting terminals.
Solution will be similar to the 2016 test question just with the length being 2l.
R = \frac{V}{I} = \frac{2l}{\,sigma \pi ab}.