Related Concepts: 2.03 Conductors, Dielectrics & Polarization Charge Densities | 2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals | 3.04 Magnetization, Magnetic Materials, Boundary Conditions & Hall Effect | 1.02 Vector Operators (Gradient, Divergence, Curl & Laplacian)

2.04 Electrostatic Boundary Conditions, Refraction Law & Poisson-Laplace Equations

Core Idea

Two closely-related high-yield blocks live here. Boundary conditions tell you how $\vec{E}$ and $\vec{D}$ behave when crossing an interface — derived by shrinking a loop (for $\vec{E}$) and a pillbox (for $\vec{D}$) onto the surface. Poisson’s and Laplace’s equations convert the postulates into a single differential equation for $V$, which is then solved to get every capacitance formula in the syllabus. Between them these topics appear in nine of the eleven papers.

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1. Electrostatic Boundary Conditions [PYQ: 2015, 2017, 2022, 2025]

Two media with permittivities ${ϵ}_1$ and ${ϵ}_2$ share a plane boundary. Take the fields close to the interface and decompose each into a tangential component (parallel to the surface) and a normal component (perpendicular to it).

[FIGURE: Interface between media 1 (${ϵ}_1$) and 2 (${ϵ}_2$) showing (a) a rectangular closed loop $abcd$ of width $\Delta w$ and height $\Delta h\to 0$ straddling the boundary, used for the tangential condition; (b) a cylindrical Gaussian pillbox of face area $\Delta s$ and height $\Delta h\to 0$, used for the normal condition. Field vectors ${\vec{E}}_1$, ${\vec{E}}_2$ shown at angles ${\alpha }_1$, ${\alpha }_2$ to the normal — source: David K. Cheng, Ch. 3, Fig. 3-23]

1.1 Tangential Component: $E_{1t}=E_{2t}$

Derivation — shrinking loop

Step 1. Apply the conservative postulate ${\oint }_C\vec{E}\cdot d\vec{l}=0$ to a small rectangular loop $abcd$ of width $\Delta w$ lying along the interface and height $\Delta h$ straddling it.

Step 2. Let $\Delta h\to 0$. The two short sides (of length $\Delta h$) contribute nothing to the line integral in this limit.

Step 3. Only the two long sides survive, traversed in opposite directions: ${\oint }_C\vec{E}\cdot d\vec{l}=E_{1t}\Delta w-E_{2t}\Delta w=0$

Step 4. Divide by $\Delta w$: $\boxed{E_{1t}=E_{2t}}\text{equivalently}\frac{D_{1t}}{{ϵ}_1}=\frac{D_{2t}}{{ϵ}_2}$

Vector form: ${\hat{a}}_{n2}\times ({\vec{E}}_1-{\vec{E}}_2)=0$

Statement: the tangential component of $\vec{E}$ is always continuous across any interface, with no exceptions.

1.2 Normal Component: $D_{1n}-D_{2n}={\rho }_s$

Derivation — shrinking pillbox

Step 1. Apply Gauss’s law ${\oint }_S\vec{D}\cdot d\vec{s}=Q_{\text{enc}}$ to a small cylindrical pillbox of face area $\Delta s$ and height $\Delta h$ straddling the interface.

Step 2. Let $\Delta h\to 0$. The flux through the curved side wall vanishes.

Step 3. Only the two flat faces contribute, with outward normals in opposite directions: ${\oint }_S\vec{D}\cdot d\vec{s}=D_{1n}\Delta s-D_{2n}\Delta s$

Step 4. The enclosed charge is the free surface charge on the interface: $Q_{\text{enc}}={\rho }_s\Delta s$

Step 5. Equate and divide by $\Delta s$: $\boxed{D_{1n}-D_{2n}={\rho }_s}\text{equivalently}{ϵ}_1{E}_{1n}-{ϵ}_2{E}_{2n}={\rho }_s$

Vector form: ${\hat{a}}_{n2}\cdot ({\vec{D}}_1-{\vec{D}}_2)={\rho }_s$

Charge-free boundary (${\rho }_s=0$, the usual dielectric–dielectric case): $D_{1n}=D_{2n}⟹{ϵ}_1{E}_{1n}={ϵ}_2{E}_{2n}$

1.3 Summary Table

Component	Condition	Charge-free case	Physical statement
Tangential $\vec{E}$	$E_{1t}=E_{2t}$	Same	${\vec{E}}_t$ is always continuous
Normal $\vec{D}$	$D_{1n}-D_{2n}={\rho }_s$	$D_{1n}=D_{2n}$	${\vec{D}}_n$ jumps by the free surface charge
Tangential $\vec{D}$	$D_{1t}/{ϵ}_1=D_{2t}/{ϵ}_2$	Same	${\vec{D}}_t$ is discontinuous unless ${ϵ}_1={ϵ}_2$
Normal $\vec{E}$	${ϵ}_1{E}_{1n}-{ϵ}_2{E}_{2n}={\rho }_s$	${ϵ}_1{E}_{1n}={ϵ}_2{E}_{2n}$	${\vec{E}}_n$ is discontinuous

Memory hook

“E is tangentially continuous, D is normally continuous.” Tangential $\to$ $\vec{E}$; normal $\to$ $\vec{D}$.

1.4 What Happens if One Medium is a Conductor? [PYQ: 2015]

PYQ — 2015 (07 marks)

Consider a plane boundary between two dielectric media (with zero conductivities) and establish a relationship between the tangential and normal components of the electric field on both sides. What happens when one of the media is a conductor?

Set medium 2 to be a perfect conductor, so ${\vec{E}}_2={\vec{D}}_2=0$ (see 2.03 Conductors, Dielectrics & Polarization Charge Densities):

General dielectric–dielectric	Becomes, with medium 2 a conductor
$E_{1t}=E_{2t}$	$E_{1t}=0$ — the field has no tangential component; it must meet the conductor perpendicularly
$D_{1n}-D_{2n}={\rho }_s$	$D_{1n}={\rho }_s$, so $E_{1n}={\rho }_s/{ϵ}_1$ — a real free surface charge appears on the conductor

In words: at a dielectric–conductor boundary the field lines become strictly normal to the conducting surface, the conductor surface becomes an equipotential, and the entire field terminates on induced free surface charge. The refraction law also degenerates — ${\alpha }_1\to 0$, meaning the field is refracted to the normal regardless of the incident angle.

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2. The Law of Refraction for Electric Field Lines [PYQ: 2015, 2017, 2019, 2022, 2025]

Derivation

Let ${\alpha }_1$, ${\alpha }_2$ be the angles that ${\vec{E}}_1$, ${\vec{E}}_2$ make with the normal to the interface. Assume a charge-free boundary (${\rho }_s=0$).

Step 1 — Resolve into components: $E_{1t}=E_1\sin {\alpha }_1,E_{1n}=E_1\cos {\alpha }_1,E_{2t}=E_2\sin {\alpha }_2,E_{2n}=E_2\cos {\alpha }_2$

Step 2 — Apply the tangential condition: E_1\sin\alpha_1 = E_2\sin\alpha_2 \tag{i}

Step 3 — Apply the normal condition: \epsilon_1 E_1\cos\alpha_1 = \epsilon_2 E_2\cos\alpha_2 \tag{ii}

Step 4 — Divide (i) by (ii): $\frac{\tan {\alpha }_1}{{ϵ}_1}=\frac{\tan {\alpha }_2}{{ϵ}_2}$

$\boxed{\frac{\tan {\alpha }_1}{\tan {\alpha }_2}=\frac{{ϵ}_1}{{ϵ}_2}=\frac{{ϵ}_{r1}}{{ϵ}_{r2}}}■$

Physical reading: field lines bend away from the normal when entering a medium of higher permittivity. If ${ϵ}_2>{ϵ}_1$, then $\tan {\alpha }_2>\tan {\alpha }_1$, so ${\alpha }_2>{\alpha }_1$.

2.1 The Standard Numerical [PYQ: 2017, 2019, 2022, 2025]

PYQ — 2017, 2019, 2022 (08/09 marks); 2025 (08 marks)

Two dielectric media with permittivities ${ϵ}_1$ and ${ϵ}_2$ are separated by a charge-free boundary. The electric field intensity in medium 1 at point $P_1$ has magnitude $E_1$ and makes an angle ${\alpha }_1$ with the normal. Determine the magnitude and direction of the electric field intensity at point $P_2$ in medium 2.

Standard solution method — memorise these four steps

Step 1 — Direction (the angle). From the refraction law: ${\alpha }_2={\tan }^{-1}\left(\frac{{ϵ}_2}{{ϵ}_1}\tan {\alpha }_1\right)$

Step 2 — Tangential component carries over unchanged: $E_{2t}=E_{1t}=E_1\sin {\alpha }_1$

Step 3 — Normal component scales by the permittivity ratio: $E_{2n}=\frac{{ϵ}_1}{{ϵ}_2}{E}_{1n}=\frac{{ϵ}_1}{{ϵ}_2}{E}_1\cos {\alpha }_1$

Step 4 — Magnitude by Pythagoras: $E_2=\sqrt{E_{2t}^2+E_{2n}^2}=E_1\sqrt{{\sin }^2{\alpha }_1+{\left(\frac{{ϵ}_1}{{ϵ}_2}\right)}^2{\cos }^2{\alpha }_1}$

Answer format: ”${\vec{E}}_2$ has magnitude $E_2$ and makes an angle ${\alpha }_2$ with the normal to the interface.”

The trap in this numerical

The question says “magnitude and direction” — two separate marks. Many students compute ${\alpha }_2$ and stop, or compute $E_2$ and stop. Always give both, and always state whether your angle is measured from the normal or from the interface — the 2017/2023 magnetic version of this question deliberately switches between the two.

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3. Poisson’s and Laplace’s Equations [PYQ: 2015, 2016, 2017, 2022, 2023, 2025]

Heavily tested — six papers

2015 (deduce both), 2016 (write both + use Laplace for capacitance), 2017 & 2023 (derive Poisson + its solution), 2022 & 2025 (derive Poisson).

Derivation

Step 1. Start from Gauss’s postulate in a material medium: $\nabla \cdot \vec{D}={\rho }_v$

Step 2. Substitute the constitutive relation $\vec{D}=ϵ\vec{E}$: $\nabla \cdot (ϵ\vec{E})={\rho }_v$

Step 3. For a homogeneous medium, $ϵ$ is constant and comes outside the divergence: $ϵ(\nabla \cdot \vec{E})={\rho }_v⟹\nabla \cdot \vec{E}=\frac{{\rho }_v}{ϵ}$

Step 4. Substitute $\vec{E}=-\nabla V$: $\nabla \cdot (-\nabla V)=\frac{{\rho }_v}{ϵ}⟹-{\nabla }^{2}V=\frac{{\rho }_v}{ϵ}$

$\boxed{{\nabla }^{2}V=-\frac{{\rho }_v}{ϵ}}\text{Poisson’s Equation}■$

Step 5. In a charge-free region (${\rho }_v=0$): $\boxed{{\nabla }^{2}V=0}\text{Laplace’s Equation}■$

Where the "homogeneous medium" assumption is used

Step 3 requires $ϵ$ to be constant so it can be pulled out of the divergence. If $ϵ=f(x,y,z)$ (an inhomogeneous medium — see 1.04 Media Properties & Conductor-Insulator Behaviour), you must keep $\nabla \cdot (ϵ\nabla V)$ intact and the simple Poisson form fails. State this assumption — it is a marked point.

3.1 The Solution of Poisson’s Equation [PYQ: 2017, 2023]

PYQ — 2017, 2023 (08/11 marks)

Derive Poisson’s equation with respect to an electric potential. What will be the solution of it?

The general (particular) solution, obtained by superposing the point-charge potential over the whole source distribution, is:

$\boxed{V=\frac{1}{4\pi ϵ}{\int }_{V^{\prime }}\frac{{\rho }_v}{R}d{v}^{\prime }}$

with the corresponding forms for surface and line sources: $V=\frac{1}{4\pi ϵ}{\int }_{S^{\prime }}\frac{{\rho }_s}{R}d{s}^{\prime },V=\frac{1}{4\pi ϵ}{\int }_{L^{\prime }}\frac{{\rho }_l}{R}d{l}^{\prime }$

where $R$ is the distance from the source element to the observation point. To this particular solution one adds any solution of the homogeneous equation ${\nabla }^{2}V=0$ needed to satisfy the boundary conditions.

Why Laplace's equation matters more in practice

Most exam problems place all the charge on conductor surfaces, leaving the region between them charge-free. There you solve ${\nabla }^{2}V=0$ subject to the electrode potentials — a boundary-value problem. Every capacitance derivation below is exactly this.

3.2 The Standard Solution Procedure

graph TD
    A["Identify the symmetry:<br/>V varies with only ONE coordinate"] --> B["Reduce ∇²V = 0 to an ODE"]
    B --> C["Integrate twice → two constants C1, C2"]
    C --> D["Apply the two electrode boundary conditions"]
    D --> E["Get V as a function of position"]
    E --> F["E = −∇V"]
    F --> G["rho_s = D_n = ε E_n at the plate"]
    G --> H["Q = rho_s × A, then C = Q / V0"]

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4. Application 1: Parallel Plate Capacitor [PYQ: 2016, 2018, 2019, 2021, 2022, 2023, 2024, 2025]

The single most-repeated numerical derivation in Chapter 2 — eight years

Asked as “determine the capacitance” (2018, 2019, 2022, 2023, 2024), “using Laplace’s equation find the capacitance” (2016), “estimate the potential and surface charge density” (2021), and “determine the surface charge density on each plate” (2022, 2025).

Master Derivation

Setup. Two large parallel plates separated by distance $d$, filled with a dielectric of permittivity $ϵ$, plate area $A$. Lower plate at $y=0$ held at $V=0$; upper plate at $y=d$ held at $V=V_0$. Fringing at the edges is neglected.

Step 1 — Reduce Laplace’s equation. By symmetry $V$ varies only with $y$: ${\nabla }^{2}V=\frac{d^{2}V}{d{y}^2}=0$

Step 2 — Integrate twice: $\frac{dV}{dy}=C_1⟹V(y)=C_{1}y+C_2$

Step 3 — Apply boundary conditions:

• At $y=0$: $V(0)=0⟹C_2=0$
• At $y=d$: $V(d)=V_0⟹C_{1}d=V_0⟹C_1=V_0/d$

Step 4 — Potential distribution (this alone answers the 2021 part (i)): $\boxed{V(y)=\frac{V_0}{d}y\text{(V)}}$ The potential varies linearly between the plates.

Step 5 — Electric field: $\vec{E}=-\nabla V=-{\hat{a}}_y\frac{dV}{dy}=-{\hat{a}}_y\frac{V_0}{d}\text{(V/m)}$ Uniform, and directed from the high-potential plate to the low-potential plate.

Step 6 — Surface charge density (this answers 2021 part (ii), 2022 and 2025). On the upper plate at $y=d$, the outward normal into the dielectric is $-{\hat{a}}_y$: $\boxed{{\rho }_s=D_n=ϵE_n=\frac{ϵV_0}{d}{\text{(C/m}}^2\text{)}}$ On the lower plate the density is $-ϵV_0/d$ — equal and opposite.

Step 7 — Total charge: $Q={\rho }_{s}A=\frac{ϵA{V}_0}{d}\text{(C)}$

Step 8 — Capacitance: $\boxed{C=\frac{Q}{V_0}=\frac{ϵA}{d}\text{(F)}}■$

Reading the result

$C\propto A$ (bigger plates hold more charge at the same voltage), $C\propto 1/d$ (closer plates produce a stronger field for the same voltage), and $C\propto ϵ$ (the dielectric polarizes and partially cancels the field, letting more charge sit at the same voltage — see 2.03 Conductors, Dielectrics & Polarization Charge Densities).

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5. Application 2: Cylindrical (Coaxial) Capacitor [PYQ: 2015, 2017]

PYQ — 2015 (10 marks), 2017 (13 marks)

A cylindrical capacitor consists of an inner conductor of radius $a$ and an outer conductor of inner radius $b$. The space between is filled with a dielectric of permittivity $ϵ$, and the length is $L$. Determine the capacitance.

Master Derivation

Setup. Inner conductor radius $a$ at potential $V_0$; outer conductor inner radius $b$ at potential $0$; length $L\gg b$ so fringing is negligible.

Step 1 — Laplace’s equation in cylindrical coordinates. $V$ varies only with $r$: ${\nabla }^{2}V=\frac{1}{r}\frac{d}{dr}\left(r\frac{dV}{dr}\right)=0$

Step 2 — Integrate once: $r\frac{dV}{dr}=C_1⟹\frac{dV}{dr}=\frac{C_1}{r}$

Step 3 — Integrate again: $V(r)=C_1\ln r+C_2$

Step 4 — Apply boundary conditions:

• At $r=b$: $V(b)=0⟹C_2=-C_1\ln b$, so $V(r)=C_1\ln (r/b)$
• At $r=a$: $V(a)=V_0⟹V_0=C_1\ln (a/b)=-C_1\ln (b/a)$ $⟹C_1=-\frac{V_0}{\ln (b/a)}$

Step 5 — Potential distribution: $V(r)=V_0\frac{\ln (b/r)}{\ln (b/a)}$ Note this is logarithmic, not linear — the cylindrical geometry concentrates the field near the inner conductor.

Step 6 — Electric field: $\vec{E}=-{\hat{a}}_r\frac{dV}{dr}={\hat{a}}_r\frac{V_0}{r\ln (b/a)}\text{(V/m)}$

Step 7 — Surface charge density on the inner conductor ($r=a$, outward normal $+{\hat{a}}_r$): ${\rho }_s=ϵE_r(a)=\frac{ϵV_0}{a\ln (b/a)}$

Step 8 — Total charge on the inner cylinder of area $A=2\pi aL$: $Q={\rho }_{s}A=\frac{ϵV_0}{a\ln (b/a)}(2\pi aL)=\frac{2\pi ϵL{V}_0}{\ln (b/a)}$

Step 9 — Capacitance: $\boxed{C=\frac{Q}{V_0}=\frac{2\pi ϵL}{\ln (b/a)}\text{(F)}}■$

Per unit length: $C^{\prime }=\frac{2\pi ϵ}{\ln (b/a)}$ (F/m) — the standard coaxial cable result.

Common slip

$\ln (b/a)$, not $\ln (a/b)$. Since $b>a$, $\ln (b/a)>0$ and the capacitance comes out positive. A negative capacitance means you inverted the ratio.

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6. Application 3: Electrohydrodynamic Pump (Poisson with ${\rho }_v\ne 0$) [PYQ: 2024]

PYQ — 2024 (13 marks)

In an electrohydrodynamic pump, the region between two electrodes is filled with a uniform charge density ${\rho }_0$. If the left electrode has potential $V_0$ and the right electrode has potential 0 V, determine the expressions for electric potential and electric field intensity at any point between the electrodes.

Solution — this one needs Poisson, not Laplace

Setup. Left electrode at $x=0$ with $V=V_0$; right electrode at $x=d$ with $V=0$; the region between carries uniform ${\rho }_v={\rho }_0$.

Step 1 — Poisson’s equation in 1D: $\frac{d^{2}V}{d{x}^2}=-\frac{{\rho }_0}{ϵ}$

Step 2 — Integrate once: $\frac{dV}{dx}=-\frac{{\rho }_0}{ϵ}x+C_1$

Step 3 — Integrate again: $V(x)=-\frac{{\rho }_0}{2ϵ}{x}^2+C_{1}x+C_2$

Step 4 — Boundary conditions:

• At $x=0$: $V(0)=V_0⟹C_2=V_0$
• At $x=d$: $-\frac{{\rho }_0{d}^2}{2ϵ}+C_{1}d+V_0=0⟹C_1=\frac{{\rho }_{0}d}{2ϵ}-\frac{V_0}{d}$

Step 5 — Potential distribution: $V(x)=-\frac{{\rho }_0}{2ϵ}{x}^2+\left(\frac{{\rho }_{0}d}{2ϵ}-\frac{V_0}{d}\right)x+V_0$ $\boxed{V(x)=V_0\left(1-\frac{x}{d}\right)+\frac{{\rho }_0}{2ϵ}x(d-x)}$

Step 6 — Electric field: $E_x=-\frac{dV}{dx}=-\left[-\frac{{\rho }_0}{ϵ}x+\frac{{\rho }_{0}d}{2ϵ}-\frac{V_0}{d}\right]$ $\boxed{\vec{E}(x)={\hat{a}}_x\left[\frac{V_0}{d}+\frac{{\rho }_0}{ϵ}\left(x-\frac{d}{2}\right)\right]\text{(V/m)}}$

Sanity checks worth writing down

• Setting ${\rho }_0=0$ recovers the empty parallel-plate result: $V=V_0(1-x/d)$, $\vec{E}={\hat{a}}_x{V}_0/d$. ✓
• The second term $\frac{{\rho }_0}{2ϵ}x(d-x)$ is a parabola vanishing at both electrodes — the space charge bows the potential without violating the fixed electrode values. ✓
• The field is now non-uniform, varying linearly with $x$ — that is exactly what drives the pumping action.

Laplace vs Poisson — pick the right one

If the region between the electrodes is charge-free, use ${\nabla }^{2}V=0$ (Laplace). If it contains a volume charge density, you must use ${\nabla }^{2}V=-{\rho }_v/ϵ$ (Poisson). The 2024 question is deliberately the Poisson case; students who default to Laplace lose the entire question.

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7. PYQ Coverage for This Note

Question (verbatim, condensed)	Marks	Year(s)
Plane boundary between two dielectrics — relate tangential and normal components. What happens when one medium is a conductor?	07	2015
Two dielectric media ${ϵ}_1$, ${ϵ}_2$, charge-free boundary, $E_1$ at angle ${\alpha }_1$ — determine magnitude and direction of ${\vec{E}}_2$.	08/09	2017, 2019, 2022
…Derive the boundary conditions and determine the magnitude and direction of the electric field in medium 2.	08	2025
Deduce the equations of Poisson’s and Laplace’s expressing the space rate of variation of electric field component.	08	2015
Write down the Laplace and Poisson’s equation. Using Laplace equation find out the capacitance of a parallel plate capacitor.	10	2016
Derive Poisson’s equation with respect to an electric potential. What will be the solution of it?	08/11	2017, 2023
Derive Poisson’s equation for electrostatics.	05/08	2022, 2025
Parallel plate capacitor, separation $d$, dielectric $ϵ$, area $A$ — determine the capacitance.	12/13	2018, 2019, 2022, 2023, 2024
Plates at 0 and $V_0$ — estimate (i) potential at any point between the plates, (ii) surface charge density on the plates.	11	2021
Fixed voltage $V_0$ across a parallel plate capacitor — determine the surface charge density on each plate.	11/12	2022, 2025
Cylindrical capacitor, inner radius $a$, outer inner radius $b$, dielectric $ϵ$, length $L$ — determine the capacitance.	10/13	2015, 2017
Electrohydrodynamic pump with uniform ${\rho }_0$ — determine $V$ and $\vec{E}$ at any point between the electrodes.	13	2024

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8. Exam Hacks & Traps

Key Exam Checkpoints

1. Boundary conditions are derived, not quoted. Draw the loop for the tangential condition and the pillbox for the normal condition, and say $\Delta h\to 0$ explicitly.
2. “E tangentially continuous, D normally continuous.” Never write $E_{1n}=E_{2n}$ — that is only true if ${ϵ}_1={ϵ}_2$.
3. State whether your angle is from the normal or from the interface. Papers switch between the two conventions deliberately; a correct number measured from the wrong reference scores zero.
4. Give both magnitude AND direction when the refraction numerical asks for it. Two separate marks.
5. State the “homogeneous medium” assumption when pulling $ϵ$ out of the divergence in the Poisson derivation.
6. Charge-free region $\to$ Laplace. Charge present $\to$ Poisson. Reading the question for this distinction is worth 13 marks in 2024.
7. Follow the fixed capacitance chain: ${\nabla }^{2}V=0\to V(x)\to \vec{E}=-\nabla V\to {\rho }_s=ϵE_n\to Q={\rho }_{s}A\to C=Q/V_0$. Do not shortcut to $C=ϵA/d$ when the question says “using Laplace’s equation”.
8. $\ln (b/a)$ with $b>a$ for the coaxial result.
9. Always state “fringing effects are neglected” — the question usually says so, and repeating it earns the assumption mark.

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9. Self-Check

1. Derive both electrostatic boundary conditions from the postulates.
2. What happens to each condition when medium 2 becomes a perfect conductor?
3. Derive $\tan {\alpha }_1/\tan {\alpha }_2={ϵ}_1/{ϵ}_2$.
4. Given $E_1$, ${\alpha }_1$, ${ϵ}_1$, ${ϵ}_2$, write the four steps to find $E_2$ and ${\alpha }_2$.
5. Derive Poisson’s equation and name the assumption used.
6. Derive $C=ϵA/d$ starting from Laplace’s equation.
7. Derive $C=2\pi ϵL/\ln (b/a)$.
8. Why does the electrohydrodynamic pump need Poisson rather than Laplace?

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Next: 2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals