Electrostatics

PYL101: Electromagnetics & Quantum Mechanics

Semester I, 2022-2023

Prof. Rohit Narula1

1 Department of Physics
The Indian Institute of Technology, Delhi

© 2016-2022 Rohit Narula. All rights reserved.

© 2016-2022 Rohit Narula. All rights reserved.
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References

Ï Introduction to Electrodynamics, David J. Griffiths [IEDJ]

1. Chapter II, Electrostatics

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Superposition

Ï Problem: What’s the force experienced by the test charge Q due to the source charge
distribution qi ?
Ï We sum the forces F i due to the individual charges in the source point charge distribution.
Ï i.e., we use the principle of superposition. . .
Ï . . . is exactly true because Maxwell’s equations are linear in both the sources (ρ and j ) and
the fields (E and B ). E.g., consider Gauss’ law,
ρ ρ ρ 1 +ρ 2 ρT
∇·E 1 = ϵ 1
0
and ∇·E 2 = ϵ 2 =⇒
0
∇·
|{z} (E 1 +E 2 )= ϵ0 =⇒∇·E T = ϵ0
linear operator

© 2016-2022 Rohit Narula. All rights reserved.

The Meaning of Electrostatics

Ï However, in general these forces depend not only on the position of the source
charges, but also on both their velocities and acceleration.1
Ï We sweep all these issues under the rug for now, and deal only with source charges
that are stationary, or fixed in space.
Ï . . . which is the domain of electrostatics.

1 With the added complication that EM fields travel at the speed of light c , and so the force experienced by
the test charge is due to what happened at the source charge distribution at a time ∼ d /c ago.
© 2016-2022 Rohit Narula. All rights reserved.
Coulomb’s Law

Ï Q: What’s the force on a test charge Q located at r due to a single point charge q
located at r 0 which is at rest?
Ï Empirical observations suggest that this is given by Coulomb’s Law,

1 qQ
F (r ) = ŝ
4πϵ0 s 2
where the separation vector s = r − r 0
Ï The constant ϵ0 is called the permittivity of free space, which in SI units is

C2
ϵ0 = 8.85 × 10−12
N.m2
Ï The Coulombic force is repulsive/attractive when the charges have the
same/opposite sign.

© 2016-2022 Rohit Narula. All rights reserved.

The Electric Field

Ï For several source point charges, using the principle of superposition we get,

Q X n q
i
F (r ) ≡ sˆi
4πϵ0 i s i2
Ï The total electric field E due to the source charges (at the test charge Q location r )
is defined as,
F (r )
E (r ) ≡
Q
Ï Giving us,
1 X n q
i
E (r ) ≡ 2
sˆi
4πϵ0 i s i

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Continuous Charge Distributions

Ï Q: What if the charges were described via a continuous charge distribution?

Ï In this case, we simply integrate,
Z
1 dq
E (r ) ≡ ŝ
4πϵ0 s2
Ï If the charge distribution is three-dimensional2 , the term d q can instead be written
as a product of the volume charge density ρ(r 0 ) and the volume element3 d τ0
giving us, Z
1 ρ(r 0 )d τ0
E (r ) ≡ ŝ
4πϵ0 s2

2 How about one- and two-dimensions?

3 What’s the variable that we’re integrating over -is it r , or r 0 ?
© 2016-2022 Rohit Narula. All rights reserved.
Vector Field

Ï Consider a single point charge q lying at rest at the origin O . Its electric field is given
by,
1 q
E (r ) ≡ r̂
4πϵ0 r 2
Ï This vector field of E can be represented by vectors that get shorter as you go farther
away from the origin; and they always point radially outward.

© 2016-2022 Rohit Narula. All rights reserved.

Field Lines

Ï Instead of drawing the vector field, we can also connect up the arrows, to form field
lines.
Ï The subtle difference here is that the magnitude of the field at a point is indicated by
the density of the field lines at that point.
Ï In this case, the electric field is stronger near the center where the field lines are close
together, and weaker farther out, where they are relatively far apart.
Ï Field lines can never cross each other at a point.4
4 Why?
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Field Lines Between Opposite Charges

Ï Note how the field lines point from positive to negative charges. . .
Ï They point outwards from positive charges
Ï They point into negative charges
Ï Q: Can field lines discern between r̂r and rr̂2 ?
Ï Q: Can field lines depict functions that are increasing such as r r̂ ?
© 2016-2022 Rohit Narula. All rights reserved.
The Electric Field Flux ΦE and Gauss’ Law

Ï The flux of E through a surface S ,

Z
ΦE = E ·da
S
is a measure of the number of field lines passing/threading through S . Because of the dot
product with d a , electric field lines ⊥ to the surface plane S contribute most appreciably to
ΦE .
Ï This suggests Gauss’ Law, P
I
q encl.
E ·da =
ϵ0
Ï Using the divergence theorem we can recast this into,
ρ
∇·E =
ϵ0
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Deriving Gauss’ Law from Coulomb’s Law
Ï . . . if you’re still not convinced, let’s start from Coulomb’s law for a 3-d charge
distribution, Z
1 ρ(r 0 )d τ0
E (r ) ≡ ŝ
4πϵ0 s2
and note that the r dependence is contained in s and further that,
µ¶
ŝ
∇r · 2 = 4πδ3 (s)
s
Ï Evaluating the divergence we have,
Z
1 ρ(r )
∇r · E = 4πδ3 (r − r 0 )ρ(r 0 )d τ0 =
4πϵ0 ϵ0
which is the differential form of Gauss’ Law.
Ï Applying the divergence theorem gives us the integral form of Gauss’ law,
I P
q encl.
E ·da =
ϵ0
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Canonical Example of Invoking Gauss’ Law

Ï Problem: calculate the field E due to a single point charge q using Gauss’ law.
Ï Gauss’ Law states, I P
q encl.
E ·da =
ϵ0
P
where q encl. = q and the field due to the point charge is, of course, turns out to be,

1 q
E= r̂
4πϵ0 r 2
Ï Think carefully about the assumptions you’re making!
Ï We’ve exploited the symmetry of the spherical (Gaussian) surface, and the radial
symmetry of the field due to to the charge.

© 2016-2022 Rohit Narula. All rights reserved.

E due to a solid sphere of total charge q

Ï Problem: What’s E outside a uniformly charged solid sphere of radius R and total
charge q ?
Ï Gauss’ Law tells us that the field is simply,

1 q
E= r̂ (r ≥ R)
4πϵ0 r 2
Ï A remarkable feature of this result is that E outside the sphere is exactly the same
as it would have been if all the charge had been concentrated at the center!
© 2016-2022 Rohit Narula. All rights reserved.
Applying Gauss’ Law Profitably

Ï The previous example should’ve alerted you to the fact that we required
Ï the symmetry of the source charge distribution and,
Ï a compatible symmetry of the Gaussian surface S ,
in order to make Gauss’ law useful in calculating E .
Ï Symmetry permitting, we usually exploit the fact,
Z Z
E ·da = | E | da
S S
Ï When ρ isn’t uniform (or doesn’t have the requisite symmetry), or if the Gaussian
surface is weird, it’s unlikely that we can apply the integral form Gauss’ law profitably
to evaluate E .

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Symmetries That Work with Gauss’ Law

Ï Apart from spherical symmetry, we can exploit Gauss’ Law when we have,
Ï Cylindrical symmetry. Make your Gaussian surface a coaxial cylinder
Ï Plane symmetry. Use a Gaussian pillbox that straddles the surface5

5 Both the cylindrical and plane case only precisely work for infinitely long, or large boxes.
© 2016-2022 Rohit Narula. All rights reserved.
Field due to a large sheet of charge density σ

Ï Problem: Find E due to an infinitely large sheet with uniform charge density σ.
Ï Applying Gauss’ Law, P
I
q encl. σA
E ·da = =
ϵ0 ϵ0

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Field due to a large sheet of charge density σ

Ï Symmetry tells us that E points away from the surface plane on each side of the
Gaussian surface/pillbox.
Ï Giving us,
1 σ
2A | E |= σA ⇒ E = n̂
ϵ0 2ϵ0
, where n̂ is the surface normal pointing away from the surface.
Ï Why’s the field the same regardless of how far away from the sheet we are?
Ï Q: Find the electric field E in, and around a parallel-plate capacitor problem (two
sheets with opposite charge, a distance d apart.) as HW.
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Is ∇ · E due to a uniform σ0 zero?
Thanks to ee3221192@iitd.ac.in

Ï "As you have mentioned that if the ∇ · E = 0 and ∇ × E = 0, in the static conditions,
then there is only one possibility to exist i.e. E = 0.
Ï However if we consider E due to an infinite charged plane sheet, it will satisfy:
1. static, as the charges on the sheet are stationary,
2. ∇ · E = 0,
3. ∇ × E = 0.
Ï Obviously the Helmholtz Theorem is not incorrect. So where my reasoning is wrong?"
© 2016-2022 Rohit Narula. All rights reserved.
Is ∇ · E due to a uniform σ0 zero?
Ï Good question. However, think about ∇ · E . It cannot possibly be zero. Why?
Ï The seemingly glib answer is that there is, after all, static charge σ in your problem!
Ï Perhaps the issue is that you’re not considering the field on both sides of the charged
sheet located in the plane z = 0, i.e., there is a field pointing in the −z direction as
well.
Ï Indeed the plane z = 0 is a region of extremely high divergence!
Ï As an exercise write the volume charge density ρ(r ) for an infinitesimally thin sheet
of charge carrying a uniform surface charge distribution σ in the plane z = 0.
Ï As regards to the Helmholtz theorem there is a subtlety that in order for it to be
applicable, the field in question must smoothly go to zero at infinity. In this
particular case the field remains the same even infinitely far away from the sheet,
precluding the use of Helmholtz’s theorem.
Ï Having said that, this isn’t too much of a problem though since the consideration of
an infinitely extended sheet of charge is obviously contrived, and not realisable in
practice.
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The Curl of E due to a Single Point Charge q

Ï Recall the paddlewheel analogy. Do you expect the E field due to a single stationary
point charge to have a curl?

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∇ × E due to a Static Charge Distribution

Ï Consider the field due to a single stationary point charge located at the origin,
q
E= r̂
4πϵ0 r 2
Ï We now integrate E over an arbitrary path a → b as shown above.
Ï In spherical coordinates d l = d r r̂ + r d θ θ̂ + r sin θd ϕϕ̂,
Zb Zb µ ¶
1 q 1 q q
E · dl = 2
d r = −
a a 4πϵ0 r 4πϵ0 r a r b

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∇ × E due to a Static Charge Distribution
Ï If the loop is closed, i.e., a = b we get,
I
E · dl = 0

Ï Thus using Stokes’ theorem / the fundamental theorem of gradients, and property (2) of
second derivatives (i.e., ∇ × ∇V = 0) we obtain,

∇× E = 0

Ï If we had multiple charges q i , each contributing their own field E i , we’d still get a zero via
the static form of Faraday’s law (principle of superposition),

∇××
|{z} E = ∇ × (E 1 + E 2 + . . .) = 0
linear operator

H
Ï In fact, E · d l = 0, and ∇ × E = 0 hold for any static charge distribution.

© 2016-2022 Rohit Narula. All rights reserved.

The Electric Potential
Ï We’ve just showed that for an arbitrary static charge distribution,

∇× E = 0
Ï Recall that such vector functions, e.g., E , can be equivalently expressed as the
gradient of a scalar potential as,

∇ × E = 0 ⇒ E = −∇(V + const.)
Ï Remarkably, all the information contained inside E (a vector) is contained inside the
scalar potential V .
Ï This is because the components of E are not indepedent6 ,

∂E x ∂E y ∂E z ∂E y ∂E x ∂E z
= , = , =
∂y ∂x ∂y ∂x ∂z ∂x

6 Why?
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The Electric Potential

Ï The scalar potential also obeys the superposition principle,

V = V1 + V2 + V3 + . . .
Ï Given E = −∇V , and the differential form of Gauss’ law we obtain Poisson’s
equation,
ρ
∇2V = −
ϵ0
Ï For a charge free region7 we get Laplace’s equation,

∇2V = 0

7 Shouldn’t E be trivially zero as dictated by the Helmholtz theorem?

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Calculating V from E
Ï We obtain V from E via the line integral,
Zr
:0

E = −∇V =⇒ V (r ) − 
V (r =
∞) = − E (r 0 ) · d l 0
∞
where the lower limit is ∞ since we (conveniently) assume the potential to be zero
there.8
Ï For a collection of stationary source charges described by a charge density ρ , we can
get V (r ) as,
Z
1 ρ(r 0 ) 0
V (r ) = dτ
4πϵ0 s

where s is the distance from the infinitesimal charge location r 0 ) to the point r .9

8 We can get away doing this since the potential V is defined only to within a constant. Adding any
constant will not change the value of E derived from the V .
9 Study Example 8, pg. 86 from [IEDJ].
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Potential Due to a Uniformly Charged Spherical Shell

Ï Problem: What’s the potential due to a uniformly charged spherical shell of radius
R?
Ï Since the entire charge lies solely at the surface of the sphere,
(
0 r ≤R
E (r ) = 1 q
4πϵ0 r 2 r̂ r >R
Ï Outside the sphere,
Zr Zr
1 q
V (r ) = − E · dl0 = − E (r 0 )d r 0 =
∞ ∞ 4πϵ0 r
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Potential Due to a Uniformly Charged Spherical Shell

Ï The potential inside (r < R ) the sphere is10 ,

Zr ZR Zr
1 q 1 q 1 q
V (r ) = − 02
dr 0 = − dr 0 − (0)d r 0 =
4πϵ0 ∞ r 4πϵ0 ∞ r 02 R 4πϵ0 R
Ï Note, that the potential is not zero inside the shell, even though the field is! Indeed,
V is a non-zero constant inside the shell.
10 Make sure you understand how to break up the integral R∞ d r 0 .
r
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Discontinuity in the Normal Component of E ⊥

Ï Problem: How does the ⊥ component of E change across a surface boundary?

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Discontinuity of the Normal Component of E ⊥

Ï Using Gauss’ Law, I

1 X 1
E ·da = q encl. = σA
S ϵ0 ϵ0
where A is the area of the Gaussian pillbox.
Ï We’re free to make the thickness of the pillbox "as short as we like", and thus the
sides do no contribute anything to the flux giving us,

above below σ
E⊥ − E⊥ =
ϵ0
Ï Of course, when σ = 0, E ⊥
above = E below
⊥

© 2016-2022 Rohit Narula. All rights reserved.

Continuity of the Tangential Component of E ∥

Ï Problem: How does the ∥ component of E change across a surface boundary?

Ï Which "law" do we invoke here?

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Continuity of the Tangential Component of E ∥

Ï Given that we know that for purely static charges,

I
E · dl = 0
we construct an Amperian loop as above with shrinking height ϵ giving us,

E ∥above = E ∥below

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Continuity of E ∥

Ï We may combine the continuity conditions for the ⊥, and ∥ components of E to

write,
σ
E above − E below = n̂
ϵ0

where is n̂ is the unit vector pointing in the direction normal to the bounding
surface.

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Continuity of V , and Discontinuity of ∇V

Ï What are the continuity conditions for the scalar potential V ?

Ï Since V is defined as,
ZP a
Vabove − Vbelow = − E · dl
Pb
and as we shrink the perpendicular path P AB to zero, so does the RHS and thus,
Vabove = Vbelow
Ï Are we really saying that the presence of charge has no effect on the potential?
Ï However, the gradient of V inherits the discontinuity from E as,

1
∇Vabove − ∇Vbelow = − σn̂
ϵ0
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The Work It Takes to Move a Test Charge

Ï Problem: Given a stationary set of source charges, how much work is needed in order
to move a test charge Q from points a to b ?

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The Work It Takes to Move a Test Charge

Ï Recall the relationship between work W and force F ,

Zb
W≡ F · dl
a
Ï The force F you must exert against the electrical force is F = −QE ,
Zb
W = −Q E · d l = +Q [V (b) − V (a)]
a

Ï The work required W is independent of the path11 taken from a to b . Such forces
are known as conservative in the language of mechanics.

11 How do we know?
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Assembling a Bunch of Point Charges
Ï Problem: How much work is required (by us) to assemble a distribution of point
charges brought in from ∞?
Ï The first charge q 1 that we bring from infinity requires zero work, because there is
no field that it feels.
Ï The second charge q 2 feels q 1 and the additional/incremental work needed is simply,12
µ ¶
q1
W2 = q 2 −0
4πϵ0 s 12
where s 12 is the final distance between q1 and q2 .
Ï Bringing in the third q 3 , which feels both q 1 , and q 2 , requires an
additional/incremental,
µ ¶
q1 q2
W3 = q 3 + −0
4πϵ0 s 13 4πϵ0 s 23
, and so on . . .
12 We somehow keep q (and all other source charges) fixed in place.
1
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Assembling a Bunch of Point Charges

Ï Adding all these up we get,

X
n 1 X n X n q q
i j
W= Wi =
i =1 8πϵ0 i =1 j 6=i s i j
Ï Note to avoid self-interaction between charges we stipulate the condition i 6= j in the
second sum □.
Ï We can recast the above equation as,

1X n
W= q i V (r i )
2 i =1
where V (r i ) is the potential due to all other charges at qi ’s location.

© 2016-2022 Rohit Narula. All rights reserved.

The Energy of a Continuous Charge Distribution

Ï The previous expression for total energy, W = 1 Pn

2 i =1 q i V (r i ) can be written in
integral form for a continuous charge distribution as,
Z
1
W= ρV d τ
2
Ï By using,
1. the differential form of Gauss’ Law ∇ · E = ρ/ϵ0 , and,
2. E = −∇V , and,
3. integration by parts
giving us finally an alternative expression for the stored electrostatic energy,
Z
ϵ0
W= |E |2 d τ
2 all space

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Ï Problem: Find the total energy contained in a uniformly charged spherical shell of total
charge q and R , starting from (a) its potential V , and (b) from its field E .
Ï (a) Using the formula containing V we get,
Z
1
W= σV d a
2
q
and realizing the potential is a constant V = 4πϵ0 R precisely at the shell13 , we get,

q2
W=
8πϵ0 R

Ï (b) Alternatively, we integrate over the field E explicitly, which is simply the field due to a
point charge everywhere outside the shell, and so we have,
Z Z∞ µ ¶2
ϵ0 2 ϵ0 Q q2
W= |E | d τ = r 2 sin θd θd ϕd r =
2 2 R 4πϵ0 r 2 8πϵ0 R

13 In fact, it must be continuous there according to the boundary condition derived earlier.
© 2016-2022 Rohit Narula. All rights reserved.
Where’s the Electrostatic Energy Stored?

Ï Note that the integral Z

1
W= ρV d τ
2
is a sum over the entire charge distribution, and so we had to (spatially) only
integrate over the shell (i.e., where the charge was located).
Ï While Z
1
W= |E |2 d τ
2 all space
is a sum over the electric field which extends from R to ∞.
Ï Q: Spatially, where’s the electrostatic energy stored?

© 2016-2022 Rohit Narula. All rights reserved.

Where’s the Electrostatic Energy Stored?

Ï If we stick to electrostatics, it’s equally OK to say that the energy is stored in the
charges themselves, or that it’s stored in the electric field.
Ï However, while studying radiation or more advanced theories (e.g., general relativity)
only the view that the energy is contained in the field is correct.

© 2016-2022 Rohit Narula. All rights reserved.

Can The Total Electrostatic Energy be Negative?

Ï The expression we just derived,

Z
ϵ0
W= E 2d τ
2 all space
implies that the total energy must always be positive!
Ï While,
1X n
W= q i V (r i )
2 i =1
for two equal but opposite charges the energy

1 q2
−
4πϵ0 s
which is clearly negative, so what gives?

© 2016-2022 Rohit Narula. All rights reserved.

Can The Total Electrostatic Energy be Negative?

Ï Consider that we went from a sum =⇒ integral, i.e.,

Z
1X n 1
W= q i V (r i ) =⇒ ρ(r 0 )V (r 0 )d τ0
2 i =1 2

Ï The integral form actually introduces a subtle error in our estimate for W .
Ï In the sum above, V (r i ) is the potential experienced by the point charge q i located at r i (i.e.,
it excludes the contribution of qi itself)
Ï Unfortunately, in the integral form above V (r 0 ), also contains the contribution of the charge
lying precisely at the location r 0 , i.e., ρ(r 0 ).
Ï For a continuous distribution, the amount of charge right at the point r 0 , i.e., ρ(r 0 ) is
vanishingly small, and its contribution to the potential V (r 0 ) is zero, and we can use
ϵ0 R
W = 2 all space |E |2 d τ without issue.

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Can The Total Electrostatic Energy be Negative?

Ï However, if we had (näively) applied the expression obtained from the integral form,
Z Z
1 0 0 0 ϵ0
W= ρ(r )V (r )d τ = |E |2 d τ
2 2 all space
for say, a single point charge q , it would blow up as,
Zµ ¶ Z∞
ϵ0 q2 2 q2 1
W= (r sin θd r d θd ϕ) = d r = +∞
2(4πϵ0 )2 r 4 8πϵ0 0 r 2
, i.e., the energy of a point charge is actually positively infinite!
Ï Bottom line: The energy of a static charge distribution can certainly be negative. In
the presence of point charges remember that one should use,

1X n
W= q i V (r i )
2 i =1

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The Lack of a Superposition Principle

Ï Claim: Because electrostatic energy is quadratic in the fields, it does not obey a
superposition principle!
Ï Indeed, Z
ϵ0
Wt ot = |(E 1 + E 2 )|2 d τ
2 Z
ϵ0
= (|E 1 |2 + |E 1 |2 + 2E 1 · E 2 )d τ
2 Z
= W1 + W2 + ϵ0 E1 · E2d τ

6= W1 + W2
Ï If we double the charge everywhere, the energy quadruples!
Ï Electrostatic energy does not obey the superposition principle.

© 2016-2022 Rohit Narula. All rights reserved.