Cylindrical Capacitors and Electric Fields

2012 Test 2 Q3

• Problem setup:
  • Cylindrical coordinates are used.
  • Assume a current I_0.
  • Find current density J, electric field E, and potential difference V_{ab}.
  • Assume a potential V_{0h}.
  • Find E, J, and current I.
  • Geometry: Cylinder with inner radius a, outer radius b, and height h. Current flows along the z-axis.
  • Resistance is denoted as R.

Calculating J and E

• 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}.

Calculating Potential Difference V

• 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}).

Calculating Resistance R

• 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)}.

Using Given V0h to Find E, J and I

• 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

• 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, E

• 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 Density, J

• Current density is given by Ohm's law: J = \,sigma E.
• J = \frac{\,sigma (V0 - Vh)}{h} \hat{z} (A/m^2).

Current, I

• 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}.

2017 Test 1 Q3

• 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.

Electric Field Intensity E(r)

• Using Gauss's law: \ointS E \cdot dS = \frac{Q{incl}}{\epsilon_0}.
• Gaussian surface: Cylinder with radius r and height h.

Region r < a

• E = 0 V/m (no enclosed charge).

Region a < 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.

Region r > b

• 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.

Potential Difference Vab

• 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})).

Line Charge Density ρl

• 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.

2016 Test 1 Q2

• 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.

Current Density J(z)

• 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.

Potential Difference

• 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}.

DC Resistance

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

2018 Test 2 Q1 (Similar to 2016 Test 1 Q2)

• 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}.