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 $b$:
  $V = - \int<em>a^b E \cdot dl = - \int</em>a^b \frac{I<em>0}{2 \pi \,sigma r h} dr \hat{r} \cdot \hat{r} = - \frac{I</em>0}{2 \pi \,sigma h} \int_a^b \frac{1}{r} dr$.
• Evaluating the integral:
  $V = - \frac{I<em>0}{2 \pi \,sigma h} [\ln(r)]</em>a^b = - \frac{I<em>0}{2 \pi \,sigma h} (\ln(b) - \ln(a)) = \frac{I</em>0}{2 \pi \,sigma h} \ln(\frac{a}{b})$.

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:
  • $V(z = 0) = V<em>0$. This gives us $K</em>2 = V_0$.
  • $V(z = h) = V<em>h$. This gives us $K</em>1 + K<em>2 = V</em>h$.
• The general solution for $V(z)$ is $V(z) = K<em>1 z + K</em>2$. Using the boundary conditions:
  $V(z) = \frac{V<em>h - V</em>0}{h}z + V_0$.

Electric Field, E

• Electric field is the negative gradient of the potential: $E = - \nabla V = - \frac{\partial V}{\partial z} \hat{z}$.
• $E = - \frac{V<em>h - V</em>0}{h} \hat{z} = \frac{V<em>0 - V</em>h}{h} \hat{z}$ (V/m).

Current Density, J

• Current density is given by Ohm's law: $J = \,sigma E$.
• $J = \frac{\,sigma (V<em>0 - V</em>h)}{h} \hat{z}$ (A/m$^2$).

Current, I

• Current is the integral of current density over the cross-sectional area:
  $I = \int J \cdot dS = \int<em>0^{2\pi} \int</em>a^b \frac{\,sigma (V<em>0 - V</em>h)}{h} r dr d\phi \hat{z} \cdot \hat{z}$.
• $I = \frac{\,sigma (V<em>0 - V</em>h)}{h} \int<em>0^{2\pi} d\phi \int</em>a^b r dr = \frac{\,sigma (V<em>0 - V</em>h)}{h} (2 \pi) [\frac{1}{2} r^2]_a^b$.
• $I = \frac{\,sigma \pi (V<em>0 - V</em>h) (b^2 - a^2)}{h}$.

2017 Test 1 Q3

• Problem setup: Infinitely long hollow cylindrical volume with uniform volume charge density $\,rho<em>v</em>0$ between $r = a$ and $r = b$, where a < b and $\,rho<em>v</em>0$ 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: $\oint<em>S E \cdot dS = \frac{Q</em>{incl}}{\epsilon_0}$.
• 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: $Q<em>{incl} = \int</em>V \rho<em>v dV = \int</em>0^h \int<em>0^{2\pi} \int</em>a^r \rho<em>v</em>0 r dr d\phi dz = \rho<em>v</em>0 h \pi (r^2 - a^2)$.
• Applying Gauss's law: $E(r) 2 \pi r h = \frac{\rho<em>v</em>0 \pi h (r^2 - a^2)}{\epsilon_0}$.
• $E(r) = \frac{\rho<em>v</em>0 (r^2 - a^2)}{2 \epsilon_0 r} \hat{r}$ V/m.

Region r > b

• Enclosed charge: $Q<em>{incl} = \int</em>V \rho<em>v dV = \int</em>0^h \int<em>0^{2\pi} \int</em>a^b \rho<em>v</em>0 r dr d\phi dz = \rho<em>v</em>0 h \pi (b^2 - a^2)$.
• Applying Gauss's law:
  $E(r) 2 \pi r h = \frac{\rho<em>v</em>0 \pi h (b^2 - a^2)}{\epsilon_0}$.
• $E(r) = \frac{\rho<em>v</em>0 (b^2 - a^2)}{2 \epsilon_0 r} \hat{r}$ V/m.

Potential Difference Vab

• $V<em>{ab} = - \int</em>a^b E \cdot dl = - \int<em>a^b \frac{\rho</em>v<em>0 (r^2 - a^2)}{2 \epsilon</em>0 r} dr \hat{r} \cdot \hat{r} = - \frac{\rho<em>v</em>0}{2 \epsilon<em>0} \int</em>a^b (r - \frac{a^2}{r}) dr$.
• $V<em>{ab} = - \frac{\rho</em>v<em>0}{2 \epsilon</em>0} [\frac{r^2}{2} - a^2 \ln(r)]<em>a^b = \frac{\rho</em>v<em>0}{2 \epsilon</em>0} [\frac{a^2}{2} - a^2 \ln(a) - \frac{b^2}{2} + a^2 \ln(b)] = \frac{\rho<em>v</em>0}{2 \epsilon_0} (\frac{a^2 - b^2}{2} + a^2 \ln(\frac{b}{a}))$.

Line Charge Density ρl

• To ensure $E(r) = 0$ for r > b, the total enclosed charge must be zero.
• $Q<em>{incl} = Q</em>v + Q<em>l = 0$, where $Q</em>v$ is the volume charge and $Q_l$ is the line charge.
• $Q<em>v = \rho</em>v_0 \pi (b^2 - a^2) h$.
• $Q<em>l = \rho</em>l h$.
• Therefore, $\,rho<em>l h + \rho</em>v_0 \pi (b^2 - a^2) h = 0$.
• $\,rho<em>l = - \rho</em>v_0 \pi (b^2 - a^2)$ 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.
• $R = \frac{V}{I} = \frac{2l}{\,sigma \pi ab}$.