Related Concepts: 2.02 Electric Potential, Equipotential Contours & Dipole Derivations | 2.04 Electrostatic Boundary Conditions, Refraction Law & Poisson-Laplace Equations | 2.01 Fundamental Postulates of Electrostatics & Gauss_s Law Applications

2.05 Electrostatic Energy, Work Done & Solved PYQ Numericals

Core Idea

Assembling a set of charges takes work, and that work is stored in the field as electrostatic energy. The derivation of $W_e=\frac{1}{2}\sum q_i{V}_i$ has been asked in 2018, 2022, 2023, 2024 and 2025 — five consecutive-era papers — and its numerical counterpart in 2019, 2020, 2023. This note also collects every remaining Chapter 2 numerical in fully worked form.

Why this note exists

Electrostatic energy is one of the highest-frequency topics in the paper (7 exam years) but had no coverage anywhere in the Chapter 2 atomic notes, and several repeated numericals had no worked solution. Both gaps are closed here.

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1. Electrostatic Energy [PYQ: 2018, 2019, 2020, 2022, 2023, 2024, 2025]

Abstract

The electrostatic energy $W_e$ of a system of charges is the total work done by an external agent in assembling that configuration by bringing each charge, one at a time and without acceleration, from infinity to its final position. Since electrostatic forces are conservative, this work is fully recoverable and is regarded as stored in the electric field.

1.1 Master Derivation: Assembling $n$ Point Charges [PYQ: 2018, 2022, 2023, 2024, 2025]

PYQ — 2018, 2022 (12), 2023 (12), 2024 (12), 2025 (13 marks)

Define electrostatic potential energy. Derive the expression for the work done in assembling $n$ (or $k$) point charges one after another from infinity.

Step-by-step derivation

Start with empty space and bring in the charges one at a time.

Step 1 — Bring in $q_1$. Empty space exerts no force, so: $W_1=0$

Step 2 — Bring in $q_2$. It must be moved through the potential $V_{21}$ established at position 2 by $q_1$: $W_2=q_2{V}_{21}=q_2\frac{q_1}{4\pi {ϵ}_0{R}_{21}}$

Step 3 — Bring in $q_3$. It moves through the combined potential of $q_1$ and $q_2$: $W_3=q_3(V_{31}+V_{32})=q_3\left(\frac{q_1}{4\pi {ϵ}_0{R}_{31}}+\frac{q_2}{4\pi {ϵ}_0{R}_{32}}\right)$

Step 4 — Continue to $q_n$, which moves through the potential of all $n-1$ charges already placed. The total work is: $W_e={\sum }_{i=1}^n{\sum }_{j<i}\frac{q_i{q}_j}{4\pi {ϵ}_0{R}_{ij}}$

Step 5 — Symmetrise. Note that assembling the charges in the reverse order gives the same total energy (the final configuration is identical). Writing $W_e$ as the average of the forward and reverse sums, every pair $(i,j)$ is now counted twice, so we divide by 2: $W_e=\frac{1}{2}{\sum }_{i=1}^n{\sum }_{\begin{array}{c}j=1\\ j\ne i\end{array}}^n\frac{q_i{q}_j}{4\pi {ϵ}_0{R}_{ij}}$

Step 6 — Recognise the inner sum as the potential $V_i$ at the location of $q_i$ due to all other charges: $V_i=\frac{1}{4\pi {ϵ}_0}{\sum }_{\begin{array}{c}j=1\\ j\ne i\end{array}}^n\frac{q_j}{R_{ij}}$

Step 7 — Final result: $\boxed{W_e=\frac{1}{2}\sum _{i=1}^n{q}_i{V}_i\text{(J)}}■$

Explain the factor of $\frac{1}{2}$ — it is a marked point

“The factor $\frac{1}{2}$ arises because the double summation counts every interacting pair twice — once as $(i,j)$ and once as $(j,i)$ — whereas each pair contributes only one physical interaction energy.”

$V_i$ excludes the charge itself

$V_i$ is the potential at $q_i$‘s location produced by every other charge, not including $q_i$‘s own field (which would be infinite for a point charge). Writing $j\ne i$ under the sum is essential.

1.2 Continuous Distributions and Energy Density

For continuous charge, the sum becomes an integral: $W_e=\frac{1}{2}{\int }_{V^{\prime }}{\rho }_{v}Vd{v}^{\prime }$

Substituting ${\rho }_v=\nabla \cdot \vec{D}$ and applying the Divergence Theorem (with the surface term vanishing at infinity) converts this into a field-only form: $W_e=\frac{1}{2}{\int }_{V^{\prime }}\vec{D}\cdot \vec{E}d{v}^{\prime }$

Electrostatic Energy Density

$w_e=\frac{1}{2}\vec{D}\cdot \vec{E}=\frac{1}{2}ϵE^2=\frac{D^2}{2ϵ}{\text{(J/m}}^3\text{)}$

Energy stored in a capacitor follows immediately: $W_e=\frac{1}{2}C{V}^2=\frac{1}{2}QV=\frac{Q^2}{2C}$

The conceptual shift worth mentioning

$W_e=\frac{1}{2}\sum q_i{V}_i$ locates the energy in the charges; $W_e=\frac{1}{2}\int \vec{D}\cdot \vec{E}dv$ locates it in the field, spread through all space. Both give the same number, but the field picture is the one that survives into time-varying electromagnetics, where energy demonstrably travels through space as a wave. This links directly to the Poynting vector in Chapter 4.

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2. Solved Numerical 1 — Energy Stored in Four Point Charges [PYQ: 2019, 2020, 2023]

PYQ — 2019, 2023 (08/10 marks)

What is electrostatic energy? What energy is stored in the field with point charges $1,-2,-3$ and $4\mu$C located on the x-axis at $x=1,2,3,4$ metres respectively?

(The 2020 variant uses charges $-2,1,3,-4\mu$C at the same positions — same method, different arithmetic.)

Solution — 2019 / 2023 version

Charges and positions:

Charge	Value ($\mu$C)	Position $x$ (m)
$q_1$	$+1$	1
$q_2$	$-2$	2
$q_3$	$-3$	3
$q_4$	$+4$	4

Step 1 — Use the pair-sum form (easier than computing four separate $V_i$): $W_e=\frac{1}{4\pi {ϵ}_0}{\sum }_{i<j}\frac{q_i{q}_j}{R_{ij}}=9\times 10^9{\sum }_{i<j}\frac{q_i{q}_j}{R_{ij}}$

Step 2 — Tabulate all six pairs. With $q$ in $\mu$C, each product $q_i{q}_j$ carries $10^{-12}$:

Pair	$q_i{q}_j$ ($10^{-12}$)	$R_{ij}$ (m)	$q_i{q}_j/R_{ij}$ ($10^{-12}$)
$(1,2)$	$(1)(-2)=-2$	1	$-2.000$
$(1,3)$	$(1)(-3)=-3$	2	$-1.500$
$(1,4)$	$(1)(4)=4$	3	$+1.333$
$(2,3)$	$(-2)(-3)=6$	1	$+6.000$
$(2,4)$	$(-2)(4)=-8$	2	$-4.000$
$(3,4)$	$(-3)(4)=-12$	1	$-12.000$
		Sum	$-12.167$

Step 3 — Multiply: $W_e=9\times 10^9\times (-12.167\times 10^{-12})=-0.1095\text{J}$

$\boxed{W_e\approx -0.110\text{J}=-109.5\text{mJ}}$

Interpreting a negative energy — worth a mark

A negative total energy means the configuration is bound: net attractive interactions dominate, and an external agent would have to supply $+0.110$ J to disassemble the system back to infinity. Say this; do not just leave a bare negative number.

Use pairs, not the $\frac{1}{2}\sum q_i{V}_i$ form, under exam pressure

The pair form has six terms; the $V_i$ form has twelve (each counted twice) plus a factor of $\frac{1}{2}$. Both give the same answer, but the pair table is faster and much harder to get wrong. Just state the equivalence: $W_e=\frac{1}{2}{\sum }_i{q}_i{V}_i={\sum }_{i<j}\frac{q_i{q}_j}{4\pi {ϵ}_0{R}_{ij}}$.

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3. Solved Numerical 2 — Work Done in a Non-Uniform Field [PYQ: 2022, 2025]

PYQ — 2022, 2025 (12 marks)

Determine the work done in carrying a $-2\mu$C charge from $P_1(2,1,-1)$ to $P_2(8,2,-1)$ in the field $\vec{E}={\hat{a}}_{x}y+{\hat{a}}_{y}x$ along the straight line joining $P_1$ and $P_2$.

Solution

Step 1 — Formula for external work: $W=-q{\int }_{P_1}^{P_2}\vec{E}\cdot d\vec{l}$

Step 2 — Form the integrand. With $d\vec{l}={\hat{a}}_{x}dx+{\hat{a}}_{y}dy+{\hat{a}}_{z}dz$ and $z$ constant ($dz=0$): $\vec{E}\cdot d\vec{l}=({\hat{a}}_{x}y+{\hat{a}}_{y}x)\cdot ({\hat{a}}_{x}dx+{\hat{a}}_{y}dy)=ydx+xdy$

Step 3 — Recognise the perfect differential. Note that $ydx+xdy=d(xy)$ so the integral is path-independent — confirming the field is conservative. (Check: $\nabla \times \vec{E}={\hat{a}}_z\left[\frac{\partial x}{\partial x}-\frac{\partial y}{\partial y}\right]={\hat{a}}_z[1-1]=0$ ✓)

Step 4 — Evaluate at the endpoints: ${\int }_{P_1}^{P_2}\vec{E}\cdot d\vec{l}=[xy{]}_{P_1}^{P_2}=(8)(2)-(2)(1)=16-2=14\text{V}$

Step 5 — Compute the work with $q=-2\mu$C $=-2\times 10^{-6}$ C: $W=-(-2\times 10^{-6})(14)=2.8\times 10^{-5}\text{J}$

$\boxed{W=28\mu \text{J}}$

The elegant shortcut — but show you earned it

Because $\vec{E}$ is conservative you never need the equation of the straight line. However, the question says “along the straight line joining $P_1$ and $P_2$”, so demonstrate that you noticed: state $\nabla \times \vec{E}=0$, conclude path independence, then use the endpoint values.

If you prefer the explicit path: the line through $(2,1)$ and $(8,2)$ is $x=6y-4$, so $dx=6dy$, and ${\int }_1^2[y(6dy)+(6y-4)dy]={\int }_1^2(12y-4)dy=[6y^2-4y{]}_1^2=(24-8)-(6-4)=14✓$

The sign trap

$W=-q\int \vec{E}\cdot d\vec{l}$ is the work done by an external agent. Here both $q$ and the sign are negative, so $W$ comes out positive — meaning you must push the negative charge against its natural tendency. Students who drop the leading minus get $-28\mu$J.

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4. Solved Numerical 3 — Spherical Cloud with Variable Density [PYQ: 2021]

PYQ — 2021 (10 marks)

A spherical uniform charge distribution in free space has ${\rho }_v=2r$ nC/m³ for $0\le r\le 10$ m and zero otherwise. Calculate $\vec{E}$ at $r=2$ m and $r=12$ m.

Read the question carefully

Despite the word “uniform”, ${\rho }_v=2r$ varies with radius. It must stay inside the integral. Writing $Q={\rho }_v\cdot \frac{4}{3}\pi r^3$ loses the entire question.

Solution

Part (a) — Inside the cloud, $r=2$ m $<10$ m

Apply Gauss’s law over a concentric sphere of radius $r$: $D_r(4\pi r^2)={\int }_0^r{\rho }_{v}d{v}^{\prime }={\int }_0^r(2r^{\prime })(4\pi r^{\prime 2}d{r}^{\prime })=8\pi {\int }_0^r{r}^{\prime 3}d{r}^{\prime }=8\pi {\left[\frac{r^{\prime 4}}{4}\right]}_0^r=2\pi r^4$

Divide through: $D_r=\frac{2\pi r^4}{4\pi r^2}=\frac{r^2}{2}{\text{nC/m}}^2$

At $r=2$ m: $D_r=\frac{2^2}{2}=2{\text{nC/m}}^2=2\times 10^{-9}{\text{C/m}}^2$ $E_r=\frac{D_r}{{ϵ}_0}=\frac{2\times 10^{-9}}{8.854\times 10^{-12}}=225.9\text{V/m}$

$\boxed{\vec{E}(2\text{m})={\hat{a}}_{r}225.9\text{V/m (radially outward)}}$

Part (b) — Outside the cloud, $r=12$ m $>10$ m

All charge is now enclosed. Integrate to the cloud radius $a=10$ m: $Q_{\text{total}}={\int }_0^{10}(2r^{\prime })(4\pi r^{\prime 2}d{r}^{\prime })=2\pi [r^{\prime 4}{]}_0^{10}=2\pi (10^4)=20000\pi \text{nC}$

$D_r=\frac{Q_{\text{total}}}{4\pi r^2}=\frac{20000\pi \times 10^{-9}}{4\pi (12{)}^2}=\frac{20000\times 10^{-9}}{576}=3.472\times 10^{-8}{\text{C/m}}^2$

$E_r=\frac{3.472\times 10^{-8}}{8.854\times 10^{-12}}=3921.4\text{V/m}$

$\boxed{\vec{E}(12\text{m})={\hat{a}}_{r}3921.4\text{V/m (radially outward)}}$

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5. Solved Numerical 4 — Field and Potential from Three Point Charges [PYQ: 2016, 2018]

PYQ — 2016, 2018 (09/10 marks)

A negative point charge of magnitude $2\mu$C is situated in air at the origin, and two positive point charges of $1\mu$C each are at $y=\pm 2$ m. Calculate the electric field strength and electric potential at a point 4 m from the origin on the x-axis.

Solution

Setup: $q_1=-2\mu$C at $(0,0,0)$; $q_2=+1\mu$C at $(0,2,0)$; $q_3=+1\mu$C at $(0,-2,0)$. Observation point $P(4,0,0)$.

Distances: $R_1=4\text{m},R_2=R_3=\sqrt{4^2+2^2}=\sqrt{20}=2\sqrt{5}\approx 4.472\text{m}$

Part (a) — Potential (scalar, so just add)

$V=\frac{1}{4\pi {ϵ}_0}\left(\frac{q_1}{R_1}+\frac{q_2}{R_2}+\frac{q_3}{R_3}\right)=9\times 10^9\left(\frac{-2\times 10^{-6}}{4}+\frac{2\times 10^{-6}}{2\sqrt{5}}\right)$ $=9\times 10^3\left(-0.5+\frac{1}{\sqrt{5}}\right)=9\times 10^3(-0.5+0.4472)=9\times 10^3(-0.0528)$ $\boxed{V=-475.2\text{V}}$

Part (b) — Field (vector — use symmetry)

By symmetry, the $y$-components from $q_2$ and $q_3$ cancel exactly; only $x$-components survive.

From $q_1$ (negative, so the field points toward it, i.e. $-{\hat{a}}_x$): $E_1=\frac{9\times 10^9\times 2\times 10^{-6}}{4^2}=1125\text{V/m}⟹{\vec{E}}_1=-{\hat{a}}_{x}1125$

From $q_2$ and $q_3$ (each at distance $\sqrt{20}$, with $\cos \theta =4/\sqrt{20}$): $E_2=E_3=\frac{9\times 10^9\times 1\times 10^{-6}}{20}=450\text{V/m}$ $E_{2x}=E_{3x}=450\times \frac{4}{\sqrt{20}}=450\times 0.8944=402.5\text{V/m}$

Wait — apply the cosine to the correct component. The $x$-component of each is $E\cos \theta$ where $\cos \theta =4/4.472=0.8944$: $E_{2x}=E_{3x}=450\times 0.8944=402.5\text{V/m}(+{\hat{a}}_x\text{ direction, since charges are positive})$

Sum: $\vec{E}={\hat{a}}_x(-1125+402.5+402.5)=-{\hat{a}}_{x}320\text{V/m}$

$\boxed{\vec{E}\approx -{\hat{a}}_{x}320\text{V/m}\text{(directed toward the origin)}}$

Note on a discrepancy in older notes

Some circulating versions of this solution report $-1044.5$ V/m, obtained by mistakenly dividing by $\sqrt{20}$ twice (once in $E=kq/R^2$ with $R^2=20$, and again when resolving the component). The correct $x$-component is $E\cos \theta =450\times (4/\sqrt{20})=402.5$ V/m, giving a net $-320$ V/m. Verify this yourself in the exam by checking that $|{\vec{E}}_2|=450$ V/m and that its $x$-component must be less than 450.

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6. Solved Numerical 5 — Potential at the Centre of a Rectangle [PYQ: 2020]

PYQ — 2020 (10 marks)

What is the potential at the centre of a rectangle whose sides are $a=2.0$ m and $b=1.0$ m, with charges $q_1=+1.0\times 10^{-8}$ C, $q_2=-2.0\times 10^{-8}$ C, $q_3=+3\times 10^{-8}$ C and $q_4=+2.0\times 10^{-8}$ C at the corners?

Solution

Step 1 — All four corners are equidistant from the centre. The half-diagonal is: $R=\frac{1}{2}\sqrt{a^2+b^2}=\frac{1}{2}\sqrt{4+1}=\frac{\sqrt{5}}{2}=1.118\text{m}$

Step 2 — Potential is a scalar, so with a common $R$ it factors out: $V=\frac{1}{4\pi {ϵ}_{0}R}{\sum }_{i=1}^4{q}_i$

Step 3 — Sum the charges: $\sum q_i=(1-2+3+2)\times 10^{-8}=4\times 10^{-8}\text{C}$

Step 4 — Evaluate: $V=\frac{9\times 10^9\times 4\times 10^{-8}}{1.118}=\frac{360}{1.118}=322.0\text{V}$

$\boxed{V\approx 322\text{V}}$

Why this one is nearly free marks

Because all four charges sit at the same distance from the centre, you never need geometry beyond the half-diagonal, and you never need vectors — potential is scalar. Spot this and the question takes ninety seconds. (Note: the electric field at the centre would require full vector addition and is a much harder question — read carefully which one is asked.)

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7. Solved Numerical 6 — $E$ and $V$ vs. Radius for a Spherical Shell [PYQ: 2025]

PYQ — 2025 (12 marks)

A positive point charge $Q$ is at the centre of a spherical conducting shell of inner radius $R_i$ and outer radius $R_o$. Illustrate the variation of electric field intensity $\vec{E}$ and electric potential $V$ as a function of radial distance $R$.

Solution — four regions

The shell is a conductor, so induced charge $-Q$ appears on its inner surface and $+Q$ on its outer surface (total charge on the shell is zero).

Region	$Q_{\text{enc}}$	$E(R)$	$V(R)$
$0<R<R_i$	$Q$	$\frac{Q}{4\pi {ϵ}_0{R}^2}$	$\frac{Q}{4\pi {ϵ}_0}\left(\frac{1}{R}-\frac{1}{R_i}+\frac{1}{R_o}\right)$
$R_i<R<R_o$ (inside conductor)	$Q+(-Q)=0$	$0$	$\frac{Q}{4\pi {ϵ}_0{R}_o}$ (constant)
$R>R_o$	$Q$	$\frac{Q}{4\pi {ϵ}_0{R}^2}$	$\frac{Q}{4\pi {ϵ}_{0}R}$

Key features to mark on the sketch:

1. $E\propto 1/R^2$ for $R<R_i$, rising steeply as $R\to 0$.
2. $E$ drops discontinuously to zero at $R=R_i$ and stays zero throughout the conductor.
3. $E$ jumps back to $Q/4\pi {ϵ}_0{R}_o^2$ at $R=R_o$ and resumes its $1/R^2$ decay.
4. $V$ is continuous everywhere — it never jumps, even where $E$ does.
5. $V$ is flat (constant) across the conductor, since $E=0$ there — the shell is an equipotential body.
6. $V\propto 1/R$ outside and rises as $1/R$ inside the cavity.

[GRAPH: Two stacked plots against $R$. Top — $E$ vs $R$: a $1/R^2$ curve from near the origin dropping to $Q/4\pi {ϵ}_0{R}_i^2$ at $R_i$, then a vertical drop to zero, a flat zero segment from $R_i$ to $R_o$, a vertical jump up to $Q/4\pi {ϵ}_0{R}_o^2$ at $R_o$, then a $1/R^2$ decay. Bottom — $V$ vs $R$: a smooth $1/R$-type curve falling from the centre, flattening into a horizontal plateau at value $Q/4\pi {ϵ}_0{R}_o$ between $R_i$ and $R_o$, then resuming a $1/R$ decay beyond $R_o$. Governing equations as tabulated above.]

The two facts the examiner is checking

$E$ can be discontinuous (it jumps at a surface charge), but $V$ is always continuous (a jump in $V$ would imply infinite $E$). And $V$ being constant inside the conductor is the direct consequence of $E=0$ there.

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8. PYQ Coverage for This Note

Question (verbatim, condensed)	Marks	Year(s)
What is electrostatic energy? Derive an equation for electrostatic energy to assemble $k$ charges one by one.	12	2018, 2022
Define electrostatic energy. Determine the electrostatic energy for assembling $k$ charges one by one.	12	2023
Define electrostatic energy. Derive the expression of electrostatic energy for a system of direct charges.	12	2024
Define electrostatic potential energy. Derive the expression for the work done in assembling $n$ point charges one after another from infinity.	13	2025
What energy is stored in the field with point charges $1,-2,-3,4\mu$C on the x-axis at $x=1,2,3,4$ m?	08/10	2019, 2023
Calculate the energy stored with charges $-2,1,3,-4\mu$C at $x=1,2,3,4$ m.	10	2020
Determine the work done in carrying a $-2\mu$C charge from $P_1(2,1,-1)$ to $P_2(8,2,-1)$ in $\vec{E}={\hat{a}}_{x}y+{\hat{a}}_{y}x$.	12	2022, 2025
Spherical distribution ${\rho }_v=2r$ nC/m³ — calculate $\vec{E}$ at $r=2$ m and $r=12$ m.	10	2021
Negative $2\mu$C at origin, two $+1\mu$C at $y=\pm 2$ — calculate $\vec{E}$ and $V$ at $x=4$ m.	09/10	2016, 2018
Positive $4\mu$C at the origin — calculate $\vec{E}$ at $z=8$ m on the z-axis.	07	2017
Potential at the centre of a rectangle $a=2.0$, $b=1.0$ with four corner charges.	10	2020
Positive $Q$ at the centre of a spherical conducting shell — illustrate $\vec{E}$ and $V$ versus $R$.	12	2025

The 2017 one-liner

Positive point charge $4\mu$C at the origin in air; find $\vec{E}$ at $z=8$ m. $E=\frac{9\times 10^9\times 4\times 10^{-6}}{8^2}=\frac{36\times 10^3}{64}=562.5\text{V/m}⟹\vec{E}={\hat{a}}_{z}562.5\text{V/m}$

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9. Exam Hacks & Traps

Key Exam Checkpoints

1. Always explain the $\frac{1}{2}$ in $W_e=\frac{1}{2}\sum q_i{V}_i$ — double-counting of pairs. It is an explicit mark.
2. $V_i$ excludes $q_i$‘s own contribution. Write $j\ne i$.
3. For numericals, use the pair form ${\sum }_{i<j}$ — six terms instead of twelve, far fewer sign errors.
4. Interpret a negative $W_e$: the configuration is bound; external work is needed to disassemble it.
5. $W=-q\int \vec{E}\cdot d\vec{l}$ for external work. Keep the leading minus. A negative charge in a positive potential difference gives positive external work.
6. Check for conservative fields. If $\nabla \times \vec{E}=0$, use endpoint values and say so — it saves several minutes.
7. Non-uniform ${\rho }_v$ stays inside the integral. The 2021 question says “uniform” but gives ${\rho }_v=2r$; trust the equation, not the adjective.
8. Potential is scalar, field is vector. Never vector-add potentials; never scalar-add fields. Exploit symmetry to cancel components before computing.
9. $E$ can jump, $V$ cannot. Essential for the 2025 shell sketch.
10. Use $1/4\pi {ϵ}_0=9\times 10^9$ — memorised, it saves time and avoids calculator slips.

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10. Self-Check

1. Derive $W_e=\frac{1}{2}\sum q_i{V}_i$ and justify the factor $\frac{1}{2}$.
2. Write the two forms of energy density and the three capacitor-energy formulas.
3. Compute the energy of $1,-2,-3,4\mu$C at $x=1,2,3,4$ m. (≈ $-0.110$ J)
4. Why is the work integral for $\vec{E}={\hat{a}}_{x}y+{\hat{a}}_{y}x$ path-independent?
5. In the 2021 cloud problem, why can’t you write $Q={\rho }_v\cdot \frac{4}{3}\pi r^3$?
6. Sketch $E$ and $V$ versus $R$ for a point charge inside a conducting shell, and state which one is continuous.

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