Potential Energy of a System of Charges

Electrostatic Potential and Capacitance · CBSE

Theory

Two charges

Assemble the configuration from infinity. Bringing $q_1$ to $\vec r_1$ costs no work — there is no field yet. Bringing $q_2$ to $\vec r_2$ costs $q_2$ times the potential there due to $q_1$ :

$U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$

Bringing them in the other order gives the same answer, because electrostatic work is path independent. $U$ characterises the configuration, not the route by which it was built.
Sign: $q_1q_2>0$ gives $U>0$ — positive external work is needed to push like charges together. $q_1q_2<0$ gives $U<0$ — positive work is needed to pull unlike charges apart, so the assembly work is negative.

Three or more charges

Bring them in one at a time. The work in the third step is $q_3$ times the potential of the first two at $\vec r_3$ , and adding the steps gives

$U=\dfrac{1}{4\pi\varepsilon_0}\left(\dfrac{q_1q_2}{r_{12}}+\dfrac{q_1q_3}{r_{13}}+\dfrac{q_2q_3}{r_{23}}\right)$

General rule: one term for every distinct pair, so $\dfrac{n(n-1)}{2}$ terms for $n$ charges. Four charges give six terms.
Work required to assemble the system $=U$ . Work required to separate the charges to infinite distance $=0-U=-U$ .

Worked patterns CBSE reuses

Equilateral triangle, side $a$ , charges $q$ , $2q$ and $-3q$ — three pairs, each at separation $a$ :

$U=\dfrac{1}{4\pi\varepsilon_0 a}\left[q(2q)+2q(-3q)+(-3q)(q)\right]=\dfrac{1}{4\pi\varepsilon_0 a}\left[2q^{2}-6q^{2}-3q^{2}\right]=\dfrac{-7q^{2}}{4\pi\varepsilon_0 a}$

Equilateral triangle, side $L$ , charges $Q$ , $q$ and $-q$ :

$U=\dfrac{kQ(-q)}{L}+\dfrac{kQq}{L}+\dfrac{k(-q)q}{L}=-\dfrac{kq^{2}}{L}$

the two $Q$ terms cancelling. The potential at $D$ , the midpoint of the side joining $-q$ and $q$ , is $V=\dfrac{2kQ}{\sqrt3\,L}$ , because $D$ is $L/2$ from each of $\pm q$ (those cancel) and $\dfrac{\sqrt3}{2}L$ from $Q$ .

Equilateral triangle, side $l$ , charges $q$ , $-4q$ , $2q$ — the two-part question:

Force on $q$ : $F_{AB}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{4q^{2}}{l^{2}}$ (attraction towards $-4q$ ) and $F_{AC}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2q^{2}}{l^{2}}$ (repulsion from $2q$ ), with $120^\circ$ between them, so

$F=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^{2}}{l^{2}}\sqrt{16+4+2\times4\times2\times\left(-\dfrac12\right)}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^{2}}{l^{2}}\left(2\sqrt3\right)$

Work to separate to infinity: $W=-U=-\dfrac{1}{4\pi\varepsilon_0 l}\left[-4q^{2}-8q^{2}+2q^{2}\right]=\dfrac{10q^{2}}{4\pi\varepsilon_0 l}$ .

Square of side $a$ with $Q$ , $q$ , $Q$ , $q$ at successive corners: four adjacent pairs at $a$ and two diagonal pairs at $a\sqrt2$ :

$U=\dfrac{1}{4\pi\varepsilon_0 a}\left[4qQ+\dfrac{q^{2}}{\sqrt2}+\dfrac{Q^{2}}{\sqrt2}\right]$

and the resultant force on one $Q$ is $F=\dfrac{Q}{4\pi\varepsilon_0 a^{2}}\left[\dfrac{Q}{2}+\sqrt2\,q\right]$ along the diagonal.

Three charges $+1\ \mu\text{C}$ , $-1\ \mu\text{C}$ , $+2\ \mu\text{C}$ at the corners of an equilateral triangle of side $10\ \text{cm}$ :

$W=\dfrac{9\times10^{9}\times10^{-12}}{0.1}\left[(1)(-1)+(1)(2)+(-1)(2)\right]=9\times10^{9}\times\dfrac{-10^{-12}}{0.1}=-0.09\ \text{J}$

Placing a third charge $Q_3$ at the midpoint of $+Q_1$ and $-Q_2$ separated by $r$ :

$W=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_1Q_3}{r/2}-\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_2Q_3}{r/2}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2Q_3}{r}\left[Q_1-Q_2\right]$

and this work vanishes at the point where the potential of the pair is zero, $x=\dfrac{rQ_1}{Q_1+Q_2}$ from $Q_1$ .

Three charges $-q$ , $Q$ , $-q$ equally spaced by $x$ on a line, with $U=0$ :

$\dfrac{-2kqQ}{x}+\dfrac{kq^{2}}{2x}=0\ \Rightarrow\ q=4Q\ \Rightarrow\ Q:q=1:4$

Work done by the field versus change in potential energy

Work done by the field $=-\Delta U$ . Moving $+3.0\ \text{nC}$ from $10\ \text{cm}$ to $15\ \text{cm}$ away from a fixed $+5.0\ \text{nC}$ : the magnitude is $kqQ\left(\dfrac{1}{r_1}-\dfrac{1}{r_2}\right)=4.5\times10^{-7}\ \text{J}$ , and because the charge moves away against a repulsive force the work done by the field is $-4.5\times10^{-7}\ \text{J}$ .
Moving a small positive charge inward along field lines of a positive source, from $Q$ (far) to $P$ (near): $W=q(V_Q-V_P)$ and $V_P>V_Q$ , so the work done by the field is negative.

Diagram to describe in words

Two charges $q_1$ and $q_2$ joined by a line marked $r_{12}$ , each with its position vector drawn from a common origin. For three charges, a triangle with $q_1$ , $q_2$ , $q_3$ at the vertices and $r_{12}$ , $r_{13}$ , $r_{23}$ marked along the sides, one pair term per side.

Standard results

Configuration	Potential energy
Two charges	$\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r_{12}}$
Three charges	Sum over the three pairs
$n$ charges	$\dfrac{n(n-1)}{2}$ pair terms
Equilateral triangle, $q,2q,-3q$ , side $a$	$\dfrac{-7q^{2}}{4\pi\varepsilon_0 a}$
Equilateral triangle, $Q,q,-q$ , side $L$	$-\dfrac{kq^{2}}{L}$
Work to assemble	$W=U$
Work to disperse to infinity	$W=-U$

Theory Questions Asked on This Subtopic

Real board questions. Answers are at the end of the page.

1
Figure shows the field lines due to a positive charge. Is the work done by the field in moving a small positive charge from $Q$ to $P$ positive or negative? Give reason. (A positive charge sits at the centre with field lines radiating outward in all directions; point $P$ lies closer to the charge, point $Q$ farther away, both along field lines.)
1 markCBSE Class 12 Physics — Very Short Answer · Foreign 2014, U · Section A, 1 markAnswer 1 ↓
2
Obtain an expression for electrostatic potential energy of a system of three charges $q$ , $2q$ and $-3q$ placed at the vertices of an equilateral triangle of side $a$ .
2 marksCBSE Class 12 Physics — SA · 2023 · Section B, 2 marksAnswer 2 ↓
3
Three point charges $Q$ , $q$ and $-q$ are kept at the vertices of an equilateral triangle of side $L$ as shown in figure. What is (i) the electrostatic potential energy of the arrangement, and (ii) the potential at point $D$ (midpoint of the side joining $-q$ and $q$ )?
5 marksCBSE Class 12 Physics — Long Answer · 2023 · Section E, 5 marksAnswer 3 ↓
4
Two point charges $q$ and $-q$ are located at $(0,0,-a)$ and $(0,0,a)$ respectively. (a) Depict the equipotential surfaces due to this arrangement. (b) Find the amount of work done in moving a small test charge $q_0$ from point $(l,0,0)$ to $(0,0,0)$ .
5 marksCBSE Class 12 Physics — Long Answer · AI 2020C · Section E, 5 marksAnswer 4 ↓
5
Obtain the expression for potential energy of an electric dipole placed with its axis at an angle $(\theta)$ to an external electric field $(\vec E)$ . What is the minimum value of the potential energy?
5 marksCBSE Class 12 Physics — Long Answer · AI 2019C · Section E, 5 marksAnswer 5 ↓
6
(a) Two point charges $+Q_1$ and $-Q_2$ are placed $r$ distance apart. Obtain the expression for the amount of work done to place a third charge $Q_3$ at the midpoint of the line joining the two charges. (b) At what distance from charge $+Q_1$ on the line joining the two charges (in terms of $Q_1$ , $Q_2$ and $r$ ) will this work done be zero?
3 marksCBSE Class 12 Physics — SA · 2020, Ev · Section C, 3 marksAnswer 6 ↓
7
Four point charges $Q$ , $q$ , $Q$ and $q$ are placed at the corners of a square of side $a$ as shown in the figure. Find (a) the resultant electric force on a charge $Q$ , and (b) the potential energy of this system.
3 marksCBSE Class 12 Physics — SA · 2018, An · Section C, 3 marksAnswer 7 ↓
8
(a) Three point charges $q$ , $-4q$ and $2q$ are placed at the vertices of an equilateral triangle $ABC$ of side $l$ as shown in the figure. Obtain the expression for the magnitude of the resultant electric force acting on the charge $q$ . (b) Find out the amount of work done to separate the charges at infinite distance.
3 marksCBSE Class 12 Physics — SA · 2018 · Section C, 3 marksAnswer 8 ↓
9
Three point charges $+1\,\mu\text{C}$ , $-1\,\mu\text{C}$ and $+2\,\mu\text{C}$ are initially infinite distance apart. Calculate the work done in assembling these charges at the vertices of an equilateral triangle of side 10 cm.
3 marksCBSE Class 12 Physics — SA · 2017 · Section C, 3 marksAnswer 9 ↓

Numericals

Every numerical set on this subtopic. Full working is at the end of the page.

10
A $+3.0\,\text{nC}$ charge $Q$ is initially at rest at a distance $r_1=10\,\text{cm}$ from a $+5.0\,\text{nC}$ charge $q$ fixed at the origin. The charge $Q$ is moved away from $q$ to a new position at $r_2=15\,\text{cm}$ . In this process, work done by the field is
- (a)
  $1.29\times10^{-5}\,\text{J}$
- (b)
  $3.6\times10^5\,\text{J}$
- (c)
  $-4.5\times10^{-7}\,\text{J}$
- (d)
  $4.5\times10^{-7}\,\text{J}$
1 markCBSE Class 12 Physics — Multiple Choice · Term I 2021-22 · Section A, 1 markAnswer 10 ↓

Answers & Solutions

Work the questions above first, then check yourself here. Each answer is set out the way it should be written in the paper.

1.↑ back to the question
Work done $=q(V_Q-V_P)$ , where $q$ is the small positive charge, and $V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_{\text{source}}}{r}$ is the potential due to the positive source charge. Since $r_P<r_Q$ , $V_P>V_Q$ , so $(V_Q-V_P)<0$ . Hence the work done by the field in moving the charge from $Q$ to $P$ is negative.
2.↑ back to the question
There are three pairs of charges, each separated by $a$ (equilateral triangle): $U=\dfrac{1}{4\pi\varepsilon_0}\left[\dfrac{q(2q)}{a}+\dfrac{2q(-3q)}{a}+\dfrac{(-3q)(q)}{a}\right]=\dfrac{1}{4\pi\varepsilon_0 a}\left[2q^2-6q^2-3q^2\right]=\dfrac{-7q^2}{4\pi\varepsilon_0 a}.$
3.↑ back to the question
(i) Potential energy of the arrangement (pairs $Q$ - $(-q)$ , $Q$ - $q$ , and $(-q)$ - $q$ , all separated by $L$ ): $U=\dfrac{kQ(-q)}{L}+\dfrac{kQq}{L}+\dfrac{k(-q)q}{L}=-\dfrac{kq^2}{L}$ (the $Q$ -terms cancel). (ii) $AD=\sqrt{L^2-\dfrac{L^2}{4}}=\dfrac{\sqrt3}{2}L$ (altitude from $A$ to $BC$ ). Potential at $D$ (distance $L/2$ from both $B(-q)$ and $C(q)$ , and $\dfrac{\sqrt3}{2}L$ from $A(Q)$ ): $V=\dfrac{k(-q)\times2}{L}+\dfrac{kq\times2}{L}+\dfrac{kQ\times2}{\sqrt3 L}=\dfrac{2kQ}{\sqrt3 L}$ (the $\pm q$ contributions cancel since both are at distance $L/2$ ).
4.↑ back to the question
(a) The equipotential surfaces are the same non-concentric family as for any dipole (closed surfaces symmetric about the $z$ -axis, denser near each charge), shown as nested/looped curves around the two charges. (b) As $W=q_0\Delta V$ : the charge $q_0$ moves from $(l,0,0)$ to $(0,0,0)$ , i.e. along the $x$ -axis. The $x$ -axis is the equatorial (perpendicular bisector) line of this dipole, on which the potential is zero everywhere ( $\Delta V=0$ ). So $W=0$ .
5.↑ back to the question
Torque on the dipole, $\tau=pE\sin\theta$ . Potential energy $=\int_{\theta_1}^{\theta_2}\tau\,d\theta=\int_{\theta_1}^{\theta_2}pE\sin\theta\,d\theta=pE[\cos\theta_1-\cos\theta_2]$ . Taking $U=0$ at $\theta_1=90°$ and $\theta_2=\theta$ : $U=pE[\cos90°-\cos\theta]=-pE\cos\theta$ , i.e. $U=-\vec p\cdot\vec E$ . Potential energy is minimum ( $U_{\min}=-pE$ ) when $\theta=0°$ , i.e. dipole moment $\vec p$ parallel to $\vec E$ (stable equilibrium).
6.↑ back to the question
(a) Work done to bring $Q_3$ from infinity to the midpoint ( $r/2$ from each charge): $W=U=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_1Q_3}{r/2}-\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q_2Q_3}{r/2}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2Q_3}{r}[Q_1-Q_2]$ . (b) Let this point be at distance $x$ from $Q_1$ (so $r-x$ from $-Q_2$ ), where $W=0$ : potential at that point due to $Q_1$ equals potential due to $Q_2$ (in magnitude): $\dfrac{kQ_1}{x}=\dfrac{kQ_2}{r-x}\Rightarrow (r-x)Q_1=xQ_2 \Rightarrow rQ_1=x(Q_1+Q_2) \Rightarrow x=\dfrac{rQ_1}{Q_1+Q_2}$ .
7.↑ back to the question
(a) With $Q$ at one corner, adjacent corners carry $q$ each (distance $a$ ) and the opposite corner carries $Q$ (distance $a\sqrt2$ ). Force due to each adjacent $q$ : $F_q=\dfrac{1}{4\pi\varepsilon_0}\dfrac{qQ}{a^2}$ (two such forces, mutually perpendicular). Force due to opposite $Q$ : $F_Q=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q^2}{(a\sqrt2)^2}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q^2}{2a^2}$ (along the diagonal). Net force: $F_{\text{net}}=F_Q+\sqrt{F_q^2+F_q^2}=F_Q+F_q\sqrt2=\dfrac{Q}{4\pi\varepsilon_0 a^2}\left[\dfrac{Q}{2}+\sqrt2\,q\right]$ , directed along the diagonal. (b) $U=U_{qQ}+U_{Qq}+U_{qQ}+U_{Qq}+U_{qq}+U_{QQ}=4U_{qQ}+U_{qq}+U_{QQ}$ $=\dfrac{4qQ}{4\pi\varepsilon_0 a}+\dfrac{q^2}{4\pi\varepsilon_0(a\sqrt2)}+\dfrac{Q^2}{4\pi\varepsilon_0(a\sqrt2)}=\dfrac{1}{4\pi\varepsilon_0 a}\left[4qQ+\dfrac{q^2}{\sqrt2}+\dfrac{Q^2}{\sqrt2}\right]$ .
8.↑ back to the question
(a) Force on $q$ due to $-4q$ : $F_{AB}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q(4q)}{l^2}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{4q^2}{l^2}$ (attractive). Force on $q$ due to $2q$ : $F_{AC}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q(2q)}{l^2}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{2q^2}{l^2}$ (repulsive). Angle between $\vec F_{AB}$ and $\vec F_{AC}$ is $120°$ . Resultant: $F=\sqrt{F_{AB}^2+F_{AC}^2+2F_{AB}F_{AC}\cos120°}$ $=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{l^2}\sqrt{16+4+2\times4\times2\times(-\tfrac12)}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{l^2}\sqrt{16+4-8}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{l^2}(2\sqrt3)$ . (b) Work done $=$ change in PE of the system $=0-(U_{AB}+U_{BC}+U_{CA})$ $=-\dfrac{1}{4\pi\varepsilon_0 l}[q(-4q)+(-4q)(2q)+(q)(2q)]=-\dfrac{1}{4\pi\varepsilon_0 l}[-4q^2-8q^2+2q^2]=\dfrac{10q^2}{4\pi\varepsilon_0 l}.$
9.↑ back to the question
Let $q_1=+1\,\mu\text{C}$ at $A$ , $q_2=-1\,\mu\text{C}$ at $B$ , $q_3=+2\,\mu\text{C}$ at $C$ ; $AB=BC=CA=10\,\text{cm}=10\times10^{-2}\,\text{m}$ . $W_A=0$ (first charge, no field to work against). $W_B=\dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1q_2}{r}=9\times10^9\times\dfrac{(1\times10^{-6})(-1\times10^{-6})}{10\times10^{-2}}=-9\times10^{-2}\,\text{J}\times10^{-1}=\dfrac{-1\times10^{-12}}{10\times10^{-2}}\times9\times10^9$ $W_C=\dfrac{1}{4\pi\varepsilon_0}\left[\dfrac{q_1q_3}{r}+\dfrac{q_2q_3}{r}\right]$ Total, $W=W_A+W_B+W_C=9\times10^9\left[\dfrac{(1)(-1)+(1)(2)+(-1)(2)}{10\times10^{-2}}\times10^{-12}\right]=9\times10^9\times\dfrac{-1\times10^{-12}}{0.1}=-0.09\,\text{J}.$
10.Answer: (C)↑ back to the question
Work done by field $=-\Delta(\text{PE})=\dfrac{kqQ}{r_1}-\dfrac{kqQ}{r_2}=9\times10^9\times5\times3\times10^{-18}\left[\dfrac{100}{10}-\dfrac{100}{15}\right]=9\times15\times10^{-7}\left[\dfrac{3-2}{30}\right]$ $=\dfrac{9\times15\times10^{-7}}{30}=4.5\times10^{-7}$ ; since both charges are positive and $Q$ moves away (against the repulsive force), work done by the field is negative: $W=-4.5\times10^{-7}\,\text{J}$ .