Theory, worked examples and 77 previous-year questions with full solutions — IIT JAM, GATE, CSIR NET-JRF and JEST, 2020 onwards.
Theory
What is Electromagnetism?
Electromagnetism is a branch of physics involving the study of the
electromagnetic force: a type of physical interaction that occurs between
electrically charged particles. The electromagnetic force is carried by
electromagnetic fields composed of electric fields and magnetic fields, and
it is responsible for electromagnetic radiation such as light. It is one of
the four fundamental interactions in nature, together with the strong
interaction, the weak interaction, and gravitation. At high energy, the weak
and electromagnetic forces are unified as a single electroweak force.
Before we discuss Electrodynamics, let us first list the four fundamental
forces in order of decreasing strength:
Strong force
Electromagnetic force
Weak force
Gravitational force
Such a short list for such a large universe—this may surprise us, since
there are so many forces we have already encountered in physics class:
Friction.
The normal force that keeps you from falling through the floor.
Chemical forces that bind molecules together.
The force of impact between two colliding billiard balls.
“All these forces are electromagnetic in nature.” So, the world we see
around us is very much electromagnetic in nature.
The broad formulation of electromagnetism covers:
Electrostatics
Magnetostatics
Electrodynamics (Maxwell's equations)
Electrostatic and magnetostatic potentials
Radiation
Relativistic electrodynamics
Question: We have two groups of charges. If we set the first group
in motion, what is the effect on the second group (see figure)?
Solution: The answer is the study of Electrodynamics.
A moving charge near a bunch of stationary charges.
Coulomb's Law
The constant \(\epsilon_0 = 8.85 \times 10^{-12}\,\mathrm{C^2/(N\cdot m^2)}\)
is called the permittivity of free space. Coulomb's law gives the
force on a test charge \(Q\) due to a single point charge \(q\) at rest a
distance \(r\) away:
\[
\mathbf{F} = \frac{1}{4\pi\epsilon_0}\frac{q\,Q}{r^2}\,\hat{\mathbf{r}}
\]
The Electric Field
Consider several point charges \(q_1, q_2, \ldots, q_n\) at distances
\(r_1, r_2, \ldots, r_n\) from a test charge \(Q\). By the principle of
superposition, the total force on \(Q\) is:
\[
\begin{aligned}
\mathbf{F}
&= \mathbf{F}_1 + \mathbf{F}_2 + \cdots \\
&= \frac{1}{4\pi\varepsilon_0}
\left(\frac{q_1 Q}{r_1^2}\hat{\mathbf{r}}_1
+ \frac{q_2 Q}{r_2^2}\hat{\mathbf{r}}_2 + \cdots\right) \\
&= \frac{Q}{4\pi\varepsilon_0}
\left(\frac{q_1}{r_1^2}\hat{\mathbf{r}}_1
+ \frac{q_2}{r_2^2}\hat{\mathbf{r}}_2
+ \frac{q_3}{r_3^2}\hat{\mathbf{r}}_3 + \cdots\right) \\
&\equiv Q\,\mathbf{E},
\end{aligned}
\]
where the electric field is defined as:
\[
\mathbf{E}(\mathbf{r})
\equiv \frac{1}{4\pi\varepsilon_0}
\sum_{i=1}^{n} \frac{q_i}{r_i^2}\,\hat{\mathbf{r}}_i.
\]
For a continuous charge distribution, a volume element \(d\tau\) carries
charge \(dq\), and the field it contributes is
\(d\mathbf{E}(\mathbf{r}) = \dfrac{1}{4\pi\varepsilon_0}\dfrac{dq}{r^2}\hat{\mathbf{r}}\),
so the total field is \(\mathbf{E} = \int d\mathbf{E}\).
The charge element \(dq\) takes the form:
\[
\boxed{dq \;\to\; \lambda\,dl \;\sim\; \sigma\,dA \;\sim\; \rho\,d\tau}
\]
where \(\lambda\) is the line charge density, \(\sigma\) the surface charge
density, and \(\rho\) the volume charge density.
Note:
Coulomb's law and the principle of superposition constitute the complete
physical input for electrostatics (charges at rest). Everything else in
electrostatics is a mathematical elaboration of these two rules.
Example 1.1. Charges \(q\), \(q\), and \(-q\) are placed at the vertices of
an equilateral triangle of side \(l\), as shown in the figure. Find the
force on each charge.
Three charges at the corners of an equilateral triangle.
Solution: The magnitude of the force between any pair of charges is
\(F = \dfrac{q^2}{4\pi\varepsilon_0 l^2}\).
The forces on charge \(q\) at A due to charges \(q\) at B and \(-q\) at C are
\(\mathbf{F}_{12}\) (along BA, repulsive) and \(\mathbf{F}_{13}\) (along AC,
attractive). By the parallelogram law, the total force on the charge at A is
\(\mathbf{F}_1 = F\,\hat{\ell}_1\), where \(\hat{\ell}_1\) is a unit vector
along BC.
Similarly, the total force on charge \(q\) at B is
\(\mathbf{F}_2 = F\,\hat{\ell}_2\) (along AC), and the total force on charge
\(-q\) at C is \(\mathbf{F}_3 = \sqrt{3}\,F\,\hat{\mathbf{n}}\), where
\(\hat{\mathbf{n}}\) bisects \(\angle\mathrm{BCA}\).
It is straightforward to verify that:
\[
\mathbf{F}_1 + \mathbf{F}_2 + \mathbf{F}_3 = \mathbf{0},
\]
which follows from Newton's third law (Coulomb's law satisfies it by
construction).
Example 1.2. An electron falls through \(h = 1.5\,\mathrm{cm}\) in a
uniform field \(E = 2.0\times10^4\,\mathrm{N/C}\)
(see figure). The field is then reversed and a proton falls the same
distance. Find the time of fall in each case.
A charged particle falling in a uniform electric field.
For the proton (field reversed, \(+e\) charge falls downward):
\[
a_p = \frac{eE}{m_p},\qquad
t_p = \sqrt{\frac{2h m_p}{eE}} = 1.3\times10^{-7}\,\mathrm{s},
\]
where \(m_p = 1.67\times10^{-27}\,\mathrm{kg}\).
The heavier proton takes longer to fall—in contrast to free fall under
gravity, where fall time is independent of mass. We justified ignoring
gravity since \(a_p = eE/m_p \approx 1.9\times10^{12}\,\mathrm{m/s^2} \gg g\).
Example 1.3. Two charges \(q_1 = +10^{-8}\,\mathrm{C}\) and
\(q_2 = -10^{-8}\,\mathrm{C}\) are placed \(0.1\,\mathrm{m}\) apart. Find the
electric field at points A, B, and C in the figure.
Electric field at three points due to a dipole configuration.
Solution:
At A: Both fields point to the right:
\[
E_A = E_{1A} + E_{2A}
= 2\times\frac{9\times10^9\times10^{-8}}{(0.05)^2}
= 7.2\times10^4\,\mathrm{N/C}, \quad \text{rightward}.
\]
At C: By symmetry both fields have magnitude
\(E_{1C} = E_{2C} = 9\times10^3\,\mathrm{N/C}\). Their components along the
line joining the charges add:
\[
E_C = 2\,E_{1C}\cos\frac{\pi}{3}
= 9\times10^3\,\mathrm{N/C}, \quad \text{rightward}.
\]
Example 1.4. Four equal charges \(q\) sit at the corners of a square of side \(a\)
(see figure). Find the neutral point(s).
Neutral points of four equal charges: \(O\) and \(P_1\)–\(P_4\).
Solution: Put \(O\) at the centre, charges at \((\pm b,\pm b)\) with \(b=a/2\).
Centre. The two charges on each diagonal cancel: \(\vec{E}(O)=0\).
Side-bisectors. On the \(x\)-axis \(E_y=0\), and with \(u=x/b\), \(f(t)=t/(1+t^2)^{3/2}\),
\[
E_x=\frac{2kq}{b^2}\left[f(1+u)-f(1-u)\right].
\]
Near \(O\), \(E_x\approx -\dfrac{kq}{\sqrt2\,b^2}\,u<0\) (towards \(O\)); at the side midpoint (\(u=1\)),
\(E_x=\dfrac{4kq}{5\sqrt5\,b^2}>0\) (away from \(O\)). So \(E_x=0\) in between; numerically
\(u\simeq0.773\), i.e. \(x\simeq0.39a\) from \(O\). By symmetry this gives four points \(P_1\)–\(P_4\).
Elsewhere. Along a diagonal \(\vec{E}\) points towards \(O\) everywhere inside, and outside the
square it points outward, so there are no other zeros.
\[
\boxed{\text{Five neutral points: } O \text{ and four points at } \approx0.39a \text{ on the side-bisectors}}
\]
\(O\) is a potential minimum in the plane of the charges; \(P_1\)–\(P_4\) are saddle points.
(Three equal charges on a triangle likewise give four.)
Example 1.5. Two equal charges \(Q_1 = Q_2 = Q\) are separated by \(2a\)
(see figure). A third charge \(q\) is placed equidistant from both.
Find \(\mathbf{F}_q\).
Find \(\mathbf{E}_q\) at the midpoint.
If \(q\) is displaced by \(x \ll a\) toward \(Q_1\), find \(F_q\) and
the oscillation frequency \(\omega\).
Two equal charges with a test charge at the midpoint.
Solution:
Forces after displacement.
Forces at equilibrium.
\(Q_1 = Q_2 = Q\) and the distances are equal, so the forces cancel:
\(F_q = 0\).
Similarly \(E_1 = E_2\) in magnitude and opposite in direction, so
\(\mathbf{E}_q = \mathbf{0}\).
After displacement \(x \ll a\) toward \(Q_1\), the restoring force is:
\[
\begin{aligned}
F_q &= \frac{kQq}{(a-x)^2} - \frac{kQq}{(a+x)^2}
= kQq\,\frac{(a+x)^2-(a-x)^2}{\left(a^2-x^2\right)^2}
= kQq\,\frac{4ax}{\left(a^2-x^2\right)^2} \\
&\approx \frac{4kQq}{a^3}\,x \qquad (x \ll a).
\end{aligned}
\]
This is a restoring force (always directed back toward the midpoint), so \(q\)
undergoes simple harmonic motion:
\[
m\,\frac{d^2x}{dt^2} = -\alpha'\,x, \qquad
\alpha' = \frac{4kQq}{a^3},
\]
giving oscillation frequency:
\[
\omega = \sqrt{\frac{\alpha'}{m}}
= \left(\frac{Qq}{\pi\epsilon_0 m\,a^3}\right)^{1/2}.
\]
Example 1.6. In the charge system of the figure, \(q_1 = q_2 = q\)
and all sides are equal to \(a\). If \(F_Q = 0\), find \(Q/q\).
Charge system with \(F_Q = 0\).
Solution: The force on \(Q\) from the diagonal charge \(Q\) and the
two adjacent charges \(q\) must balance. The diagonal distance is \(a\sqrt{2}\),
so:
\[
F_Q = \frac{kQ^2}{2a^2} + \frac{2kQq}{a^2}\cdot\frac{1}{\sqrt{2}} = 0.
\]
\[
\Rightarrow \quad \frac{Q}{q} = -2\sqrt{2}.
\]
Example 1.7. A plane sheet produces a constant electric field \(E_0\).
A charged bob is attached to the sheet by a silk thread
(see figures). At equilibrium, find the angle \(\theta\)
between the thread and the sheet. When does \(\theta = 90\ensuremath{^\circ}\)?
Resolution of forces on the bob.
Pendulum near a charged plane sheet.
Solution: Balancing the tension components:
\[
T\cos\theta = mg, \qquad T\sin\theta = qE_0.
\]
\[
\Rightarrow \quad
\theta = \tan^{-1}\!\left(\frac{qE_0}{mg}\right).
\]
When \(\theta = 90\ensuremath{^\circ}\), we need \(qE_0 \gg mg\), i.e., \(E_0 \to \infty\).
Example 1.8. Two charged spheres (charge \(q\) each) are in equilibrium
as in the figure (\(\theta\) very small). The charge begins leaking.
Find the velocity as a function of separation \(x\).
Solution:
Forces after charge begins to leak.
Two charged spheres in equilibrium.
At equilibrium, using \(\sin\theta \approx \theta = x/(2l)\):
\[
\frac{kq^2}{x^2} = mg\,\frac{x}{2l}
\;\Rightarrow\;
q = \alpha\,x^{3/2}, \quad \alpha = \sqrt{\frac{mg}{2lk}}.
\]
Differentiating with respect to time:
\[
\frac{dq}{dt} = \frac{3\alpha}{2}\,x^{1/2}\,v
\;\Rightarrow\;
v(x) = \frac{1}{\frac{3\alpha}{2}\sqrt{x}}\frac{dq}{dt}
\propto \frac{1}{\sqrt{x}}.
\]
Example 1.9. A ring of radius \(r\) carries total charge \(q\) uniformly.
An additional charge \(q_0\) is placed at the centre
(see figures). Find the change in tension \(\Delta T\)
in the ring.
Force balance on a ring element.
Charged ring under tension.
Solution: Before placing \(q_0\), for a small element subtending
angle \(d\theta\): \(T\,d\theta = F_e\) (self-repulsion).
After placing \(q_0\), the charge element is
\(dq = \dfrac{q\,r\,d\theta}{2\pi r} = \dfrac{q\,d\theta}{2\pi}\), and:
\[
(T+\Delta T)\,d\theta = F_e + \frac{kq_0\,dq}{r^2}
\;\Rightarrow\;
\Delta T\,d\theta = \frac{kq_0}{r^2}\cdot\frac{q\,d\theta}{2\pi}.
\]
\[
\boxed{\Delta T = \frac{kq_0\,q}{2\pi r^2} = \frac{q_0\,q}{8\pi^2\varepsilon_0 r^2}.}
\]
Example 1.10. Two charges \(+q\) and \(+q\) are separated by distance \(d\).
A point \(P\) lies at perpendicular distance \(z\) from the midpoint
(see figure).
Find \(\mathbf{E}_P\).
Evaluate when \(z = 0\).
Evaluate when \(z \gg d\).
Electric field of a two-charge system at point \(P\).
(b) At \(z = 0\): \(\mathbf{E}_P = \mathbf{0}\) (the two fields point in
opposite directions and cancel).
(c) For \(z \gg d\):
\(\mathbf{E}_P \approx \dfrac{2kq}{z^2}\,\hat{z}\)—the pair looks like a
single charge \(2q\) at the foot of the perpendicular.
Concept Checks — Coulomb's Law, Electric Field & Gauss's Law
Concept check.
A thin conducting square loop of side \(a\) is placed in the first quadrant of the \(xy\)-plane with one vertex at origin. A changing magnetic field \(\partial \mathbf{B}/\partial t = B_0\hat{z}\) (\(B_0\) = const) is applied. The magnitude of the induced EMF in the loop is?
Show answer
\(\mathcal{E} = -d\Phi_B/dt = B_0 a^2\).
Work Done in Moving a Charge; Electric Potential Energy
If we move a test charge \(Q\) along a path element \(d\mathbf{l}\) in an
electric field \(\mathbf{E}\), the work done by an external agent
(against the field) is:
\[
dW = -Q\,\mathbf{E}\cdot d\mathbf{l}.
\]
Over a finite path from an initial to a final point:
\[
W = -Q\int_{\text{init}}^{\text{final}}\mathbf{E}\cdot d\mathbf{l}.
\]
Note: \(dW = 0\) whenever \(\mathbf{E}\perp d\mathbf{l}\)—motion
along an equipotential requires no work.
Example 1.11. The electric field at a point is
\[
\mathbf{E} = \frac{1}{z^2}
\bigl(8xyz\,\hat{x} + 4x^2z\,\hat{y} - 4x^2y\,\hat{z}\bigr).
\]
Find the work done in moving \(Q = 6\,\mathrm{nC}\) a distance
\(|d\mathbf{l}| = 2\,\mu\mathrm{m}\) from \(P(2,-2,3)\) in the direction
\(\hat{u} = -\tfrac{6}{7}\hat{x} + \tfrac{3}{7}\hat{y} + \tfrac{2}{7}\hat{z}\).
Solution: The displacement vector is
\(d\mathbf{l} = |d\mathbf{l}|\,\hat{u} = 2\times10^{-6}\hat{u}\), so the
final point is:
\[
x = 2 - \tfrac{12}{7}\times10^{-6} \approx \tfrac{2}{7},\quad
y = -2 + \tfrac{6}{7}\times10^{-6}\approx -\tfrac{8}{7},\quad
z = 3 + \tfrac{4}{7}\times10^{-6}\approx \tfrac{25}{7}.
\]
Integrating along the path segment by segment:
\[
W = -Q\left[\int_{\text{path}}\frac{8xy}{z}dx
+ \int_{\text{path}}\frac{4x^2}{z}dy
- \int_{\text{path}}\frac{4x^2y}{z^2}dz\right]
\approx -63.4\times10^{-15}\,\mathrm{J} = -63.4\,\mathrm{fJ}.
\]
(The work is negative: the field does work on the charge.)
Example 1.12. \(Q = 2\,\mathrm{C}\), \(\mathbf{E} = y\hat{x} + x\hat{y} + 2\hat{z}\).
Move from \(B(1,0,1)\) to \(A(0.8,0.6,1)\) along:
(a) the circle \(x^2+y^2 = 1\); (b) the straight line.
The potential difference between points A and B is defined as the
work done per unit charge in moving a test charge from B to A:
\[
V_{AB} = V_A - V_B = -\int_{r_B}^{r_A}\frac{kq}{r^2}\,dr
= kq\!\left[\frac{1}{r_A} - \frac{1}{r_B}\right].
\]
Choosing infinity as the reference point (\(V(\infty)=0\)):
\[
\boxed{V(r) = \frac{1}{4\pi\epsilon_0}\frac{q}{r}.}
\]
All points at the same potential form an equipotential surface.
For a positive point charge, these are concentric spheres. The electric
field \(\mathbf{E}\) is always perpendicular to equipotential surfaces and
points in the direction of decreasing potential.
Note:
The choice of reference point (where \(V = 0\)) is arbitrary, but
potential differences \(V_A - V_B\) are physically meaningful and
independent of this choice.
Concept Checks — Electric Potential
Concept check.
Equipotential surfaces are \(4x^2+4y^2+z^2=\text{const}\). The electric field at origin is:
(a) \(\vec{0}\)
(b) along \(+\hat{z}\)
(c) along \(-\hat{z}\)
(d) indeterminate
Show answer
(a). \(V=f(4x^2+4y^2+z^2)\); \(\vec{E}=-\nabla V = 0\) at origin since all partial derivatives vanish at \((0,0,0)\).
Potential Gradient and Equipotential Surfaces
The relationship between the electric field and the potential is:
\[
\mathbf{E} = -\nabla V.
\]
This can be seen from \(dV = -\mathbf{E}\cdot d\mathbf{l}\): comparing with
\(dV = \tfrac{\partial V}{\partial x}dx + \tfrac{\partial V}{\partial y}dy
+ \tfrac{\partial V}{\partial z}dz\) gives:
\[
E_x = -\frac{\partial V}{\partial x},\quad
E_y = -\frac{\partial V}{\partial y},\quad
E_z = -\frac{\partial V}{\partial z}.
\]
In general, the magnitude of \(\mathbf{E}\) equals the maximum rate of change
of \(V\) with distance, and the direction of \(\mathbf{E}\) is opposite to the
direction of steepest increase of \(V\).
(b) Setting \(\phi = \phi_0\):
\[
\frac{x^2}{\left(\sqrt{\phi_0/a}\right)^2}
+ \frac{y^2}{\left(\sqrt{\phi_0/a}\right)^2}
+ \frac{z^2}{\left(\sqrt{\phi_0/b}\right)^2} = 1.
\]
This is an ellipsoid (see figure). If \(a = b\), it
reduces to a sphere.
Ellipsoidal equipotential surface (\(a = b \neq c\)).
(a) \(\mathbf{E} = -2ax\hat{x} + 2ay\hat{y}\). The equipotentials for \(a>0\)
are rectangular hyperbolae (see figure).
(b) \(\mathbf{E} = -ay\hat{x} - ax\hat{y}\).
Equipotential surfaces for \(\varphi = a(x^2-y^2)\).
Example 1.17. Find the electrostatic energy stored in the system of four
equal charges \(q\) at the corners of a square of side \(a\)
(see figure).
Four charges at the corners of a square.
Solution: The energy equals the work to assemble the charges from
infinity (reference \(V = 0\) at infinity). Label charges 1–4 clockwise from
top-left:
Interaction energy between two charges:
\(u_{12} = \dfrac{1}{4\pi\epsilon_0}\dfrac{q_1 q_2}{r_{12}}\).
Continuous Charge Distribution Problems
Example 1.18. Find the electric field at a point \(P\) on the axis of a
uniformly charged cone (charge density \(\rho\), height \(h\), base radius \(a\)),
the figures.
Geometry for the cone integration.
Uniformly charged cone; field at apex.
Solution: By symmetry only the axial component survives.
For a volume element at \((r,\theta,\phi)\):
\[
dE = k\frac{\rho\,r^2\sin\theta\,d\theta\,d\phi\,dr}{r^2}\cos\theta
= k\rho\sin\theta\cos\theta\,d\theta\,d\phi\,dr.
\]
Integrating over \(\phi\in[0,2\pi]\) and \(r\in[0,h/\cos\theta]\):
\[
E = k\rho\cdot 2\pi\int_0^\alpha h\sin\theta\cos\theta\,\frac{h}{\cos\theta}\,d\theta
\]
\[
\boxed{\mathbf{E} = k\rho\,2\pi h^2
\!\left(1 - \frac{h}{\sqrt{R^2+h^2}}\right)(-\hat{z}).}
\]
Example 1.19. Find the electric field a distance \(z\) above one end of a
line segment of length \(2L\) carrying uniform line charge \(\lambda\)
(see figure). What is \(\mathbf{E}\) when \(z\gg L\)?
Straight line segment of length \(2L\).
Solution: By symmetry the transverse components cancel. Using
\(\displaystyle\int\frac{dx}{(x^2+a^2)^{3/2}} = \frac{x}{a^2\sqrt{x^2+a^2}}\):
\[
E_{\text{net}}
= \frac{2\lambda z}{4\pi\varepsilon_0}
\int_0^L\frac{dx}{(x^2+z^2)^{3/2}}
= \frac{1}{4\pi\varepsilon_0}\frac{2\lambda L}{z\sqrt{z^2+L^2}}.
\]
(a) For \(z \gg L\): \(E \approx \dfrac{q}{z^2}\), \(q = 2\lambda L\)
(behaves like a point charge).
(b) For \(L\to\infty\): \(E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2\lambda}{z}\)
(standard infinite wire result).
Example 1.20. Find the electric field a distance \(z\) above one end
of a line segment of length \(L\) (see figure).
Straight line segment of length \(L\) (field above one end).
Example 1.21. Find the electric field a distance \(z\) above the centre of
a square loop (side \(a\)) carrying uniform line charge \(\lambda\)
(see figure).
Square loop of side \(a\).
Solution: Use the result for a segment of length \(2L = a\) at
perpendicular distance \(\tilde{z} = \sqrt{z^2+a^2/4}\) from the field point.
Each side contributes a field \(E_1\); the four contributions add by symmetry:
\[
E_1 = \frac{1}{4\pi\varepsilon_0}
\frac{a\lambda}{\sqrt{z^2+\tfrac{a^2}{4}}}
\cdot\frac{1}{\sqrt{z^2+\tfrac{a^2}{2}}},
\]
\[
E_{\text{net}} = 4E_1\cos\theta
= \frac{1}{4\pi\varepsilon_0}
\frac{4\lambda a\,z}{\left(z^2+\tfrac{a^2}{4}\right)\sqrt{z^2+\tfrac{a^2}{2}}}.
\]
Example 1.22. Find the electric field a distance \(z\) above the centre of
a flat circular disk of radius \(R\) with surface charge \(\sigma\)
(see figure).
(a) Limit \(R\to\infty\); (b) Limit \(z\gg R\).
Flat circular disc with surface charge \(\sigma\).
Solution: Break the disk into rings of radius \(r\) and thickness
\(dr\). Each ring has effective line charge \(\lambda = \sigma\,dr\):
\[
E_{\text{disk}}
= \frac{\sigma}{2\varepsilon_0}
\!\left[1 - \frac{z}{\sqrt{R^2+z^2}}\right]\hat{z}.
\]
(b) \(z\gg R\) (expand to first order in \(R/z\)):
\[
E \approx \frac{k\pi\sigma R^2}{z^2} = \frac{\sigma\pi R^2}{4\pi\varepsilon_0 z^2}
\quad\text{(point charge $q = \sigma\pi R^2$)}. \checkmark
\]
Example 1.23. A thin ring of radius \(R\) carries linear charge density
\(\lambda = \lambda_0\cos\varphi\). Find \(\mathbf{E}\) at the centre.
Solution: The element \(dq = \lambda_0 R\cos\varphi\,d\varphi\)
contributes \(dE\cos\varphi\) along the axis of symmetry:
\[
E = \frac{\lambda_0}{4\pi\varepsilon_0 R}
\int_0^{2\pi}\cos^2\!\varphi\,d\varphi
= \frac{\lambda_0}{4\pi\varepsilon_0 R}\cdot\pi
= \frac{\lambda_0}{4\varepsilon_0 R}.
\]
(Used \(\langle\cos^2\varphi\rangle = \tfrac{1}{2}\), so the integral \(= \pi\).)
Example 1.24. For the same ring, find \(\mathbf{E}\) on the axis at
distance \(x\) from the centre. Evaluate for \(x\gg R\).
Ring with \(\lambda = \lambda_0\cos\varphi\); field on axis.
Solution: With source element at azimuthal angle \(\varphi\),
\(dq = \lambda_0 R\cos\varphi\,d\varphi\), and the displacement from source
to field point is \(\mathbf{r}-\mathbf{R} = x\hat{k} - R(\cos\varphi\,\hat{i}
+ \sin\varphi\,\hat{j})\), with \(|\mathbf{r}-\mathbf{R}| = \sqrt{R^2+x^2}\).
Integrating and using
\(\int_0^{2\pi}\cos\varphi\,d\varphi = \int_0^{2\pi}\sin\varphi\,d\varphi = 0\)
and \(\int_0^{2\pi}\cos^2\varphi\,d\varphi = \pi\):
\[
\mathbf{E} = -\frac{\lambda_0 R^2}{4\varepsilon_0(R^2+x^2)^{3/2}}\,\hat{i},
\qquad
E = \frac{\lambda_0 R^2}{4\varepsilon_0(R^2+x^2)^{3/2}}.
\]
For \(x\gg R\): \(E \approx \dfrac{p}{4\pi\varepsilon_0 x^3}\)
where \(p = \lambda_0\pi R^2\) is the effective dipole moment.
Gauss's Law
Gauss's law states that the total electric flux through any closed
surface equals the total enclosed charge divided by \(\epsilon_0\):
\[
\oint_S \mathbf{E}\cdot d\mathbf{a} = \frac{Q_{\text{enc}}}{\epsilon_0}
= \frac{1}{\epsilon_0}\int_{\mathcal{V}}\rho\,d\tau.
\]
Applying the divergence theorem to the left side:
\[
\oint_S \mathbf{E}\cdot d\mathbf{a}
= \int_{\mathcal{V}}(\nabla\cdot\mathbf{E})\,d\tau,
\]
and since this holds for any volume, the integrands must be equal
pointwise:
\[
\boxed{\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0}.}
\quad\text{(Gauss's law, differential form)}
\]
Point charge. With \(\rho=q\,\delta^3(\vec{r})\) and \(\nabla\cdot(\rhat/r^2)=4\pi\delta^3(\vec{r})\)
(formula sheet, p. ),
\[
\nabla\cdot\left(\frac{q}{4\pi\epsilon_0}\frac{\rhat}{r^2}\right)=\frac{q}{\epsilon_0}\,\delta^3(\vec{r}),
\]
zero everywhere except at the charge, the source of the entire flux \(q/\epsilon_0\).
Key remarks:
We now have Gauss's law in both integral and differential form.
Gauss's law is always true, but is only useful
when the charge distribution has sufficient symmetry.
Symmetry guides the choice of Gaussian surface:
enumerate[label=*.]
Spherical symmetry: use a concentric sphere.
Cylindrical symmetry: use a coaxial cylinder.
Planar symmetry: use a Gaussian “pillbox” that
straddles the surface.
enumerate
Example 1.25. Show that a charge \(q > 0\) cannot be in stable
electrostatic equilibrium in free space.
A closed surface \(S\) surrounding charge \(q\).
Solution: Enclose \(q\) in a small surface \(S\) (see figure).
For stable equilibrium, \(\mathbf{E}\) due to all other charges must point
inward everywhere on \(S\), giving negative flux. But Gauss's law says
the flux through \(S\) must be zero (since \(q\) itself is the only enclosed
charge, and the field of external charges has zero net flux through \(S\)).
Contradiction. Hence stable equilibrium is impossible—this is Earnshaw's
theorem.
Example 1.26. Find the electric field of an infinite plane sheet with
surface charge density \(\sigma\).
Gaussian pillbox for an infinite plane sheet.
Solution: By symmetry \(\mathbf{E}\) is perpendicular to the plane
and equal in magnitude on both sides. Use a pillbox of face area \(\Delta S\):
\[
2E\,\Delta S = \frac{\sigma\,\Delta S}{\varepsilon_0}
\;\Rightarrow\;
\boxed{E = \frac{\sigma}{2\varepsilon_0}.}
\]
The field is uniform on both sides and points away from the sheet (for
\(\sigma > 0\)).
Example 1.27. Find the field of two parallel planes with charge
densities \(+\sigma\) and \(-\sigma\) (see figure).
Two oppositely charged parallel plane sheets.
Solution: By superposition, the individual fields add between the
plates and cancel outside:
\[
E_{\text{between}} = \frac{\sigma}{\varepsilon_0}, \qquad
E_{\text{outside}} = 0.
\]
This is the ideal parallel-plate capacitor (valid near the centre, away from
edges).
Example 1.28. Find the field of an infinite cylindrical shell
(charge \(\lambda\) per unit length, radius \(a\)), the figure.
Coaxial Gaussian cylinder for an infinite cylindrical shell.
Solution: Use a coaxial cylinder of radius \(r\) and height \(h\).
The flux through the end caps is zero; through the curved surface it is
\(E\cdot 2\pi r h\).
\[
r > a:\quad E\cdot 2\pi r h = \frac{\lambda h}{\varepsilon_0}
\;\Rightarrow\;
\boxed{E = \frac{\lambda}{2\pi\varepsilon_0 r}.}
\]
\[
r < a:\quad \text{no enclosed charge} \;\Rightarrow\; E = 0.
\]
Example 1.29. Find the field of a uniformly charged spherical
shell (total charge \(q\), radius \(a\)), the figure.
Spherical shell of charge \(q\).
Solution:
\[
r > a:\quad E\cdot 4\pi r^2 = \frac{q}{\varepsilon_0}
\;\Rightarrow\; E = \frac{q}{4\pi\varepsilon_0 r^2}.
\qquad r < a:\quad E = 0.
\]
Example 1.30. Find the field of a uniformly charged solid sphere
(total charge \(q\), radius \(a\)), the figure.
Solid sphere of uniform charge density.
Solution: For \(r < a\), the enclosed charge is
\(q' = q(r/a)^3\):
\[
E_r\cdot 4\pi r^2 = \frac{q(r/a)^3}{\varepsilon_0}
\;\Rightarrow\;
E_r = \frac{q}{4\pi\varepsilon_0 a^3}\,r \quad (r \leq a).
\]
The field grows linearly inside (see figure).
\(E\) vs \(r\) for a uniformly charged solid sphere.
Example 1.31. Which of the following is an impossible electrostatic
field?
(a) \(\mathbf{E} = k(xy\hat{x} + 2yz\hat{y} + 3xz\hat{z})\);
(b) \(\mathbf{E} = k(y^2\hat{x} + (2xy+z^2)\hat{y} + 2yz\hat{z})\).
Solution: An electrostatic field must satisfy
\(\nabla\times\mathbf{E} = \mathbf{0}\).
Example 1.32. Find the potential inside and outside a uniformly charged
solid sphere (radius \(R\), total charge \(q\)). Compute \(\nabla V\) and verify
it gives the correct field.
Solution:
\[
V(r) = \begin{cases}
\dfrac{q}{4\pi\epsilon_0}\dfrac{1}{r} & r > R, \\[6pt]
\dfrac{q}{4\pi\epsilon_0}\dfrac{1}{2R}\!\left(3 - \dfrac{r^2}{R^2}\right)
& r \leq R.
\end{cases}
\]
Checking: \(\mathbf{E} = -\nabla V\) gives
\(\mathbf{E} = \dfrac{q}{4\pi\epsilon_0 r^2}\hat{r}\) for \(r>R\) and
\(\mathbf{E} = \dfrac{qr}{4\pi\epsilon_0 R^3}\hat{r}\) for \(r < R\). ✓
Example 1.33. Find the potential a distance \(s\) from an infinite
straight wire (\(\lambda\)). Verify \(\nabla V = -\mathbf{E}\).
Solution: Choose reference at \(s = a\):
\[
V(s) = -\frac{\lambda}{2\pi\varepsilon_0}\ln\!\left(\frac{s}{a}\right).
\]
Then \(-\nabla V = -\left(-\dfrac{\lambda}{2\pi\varepsilon_0 s}\right)\hat{s}
= \dfrac{\lambda}{2\pi\varepsilon_0 s}\hat{s} = \mathbf{E}\). ✓
(Note: \(a = \infty\) or \(a = 0\) cannot be used as reference since
\(\ln(\infty/\infty)\) or \(\ln(0)\) diverges.)
Example 1.34. A thick spherical shell carries charge density
\(\rho = k/r^2\) for \(a \leq r \leq b\) (see figure). Find the
potential at the centre.
Thick spherical shell with charge density \(\rho = k/r^2\).
Solution: First find \(\mathbf{E}\) by Gauss's law in all three
regions, then integrate \(V(0) = -\int_\infty^0 \mathbf{E}\cdot d\mathbf{l}\):
\[
V(0) = \frac{k}{\epsilon_0}\ln\!\left(\frac{b}{a}\right).
\]
Example 1.35. A long coaxial cable has inner cylinder (radius \(a\),
volume charge \(\rho\)) and outer shell (radius \(b\), surface charge such
that the cable is overall neutral), the figure. Find
\(V(0) - V(b)\).
Relations Between \(\rho\), \(E</h2>
\) and \(V\)
Relationships between \(\rho\), \(\mathbf{E}\), and \(V\) in
electrostatics.
The three key quantities in electrostatics—charge density \(\rho\), electric
field \(\mathbf{E}\), and potential \(V\)—are linked by:
\[
\mathbf{E} = -\nabla V, \quad
\nabla\cdot\mathbf{E} = \frac{\rho}{\epsilon_0}, \quad
\nabla^2 V = -\frac{\rho}{\epsilon_0} \;\text{(Poisson's equation)}.
\]
Concept Checks — Electrostatic Energy
Concept check.
Electrostatic potential \(V=2xyz\) V. Energy density at a point:
(a) \(2\eps(y^2z^2+x^2z^2+x^2y^2)\)
(b) \(\eps(y^2z^2+x^2z^2+x^2y^2)\)
(c) \(4\eps(y^2z^2+x^2z^2+x^2y^2)\)
(d) \(\eps(yz+xz+xy)\)
Show answer
(a). \(\vec{E}=-\nabla V = -(2yz\hat{x}+2xz\hat{y}+2xy\hat{z})\); \(u=\frac{\eps}{2}E^2 = 2\eps(y^2z^2+x^2z^2+x^2y^2)\).
Boundary Conditions
At a surface carrying surface charge density \(\sigma\), the electric field
changes abruptly. Let \(\hat{\mathbf{n}}\) point from region below to
region above the surface.
Derivation: Normal Component
Apply Gauss's law to a thin pillbox (face area \(A\), height \(\epsilon\to 0\))
straddling the surface (see figure). The curved side contributes
nothing:
\[
E_{\text{above}}^\perp A - E_{\text{below}}^\perp A
= \frac{\sigma A}{\epsilon_0}.
\]
\[
\boxed{E_{\text{above}}^\perp - E_{\text{below}}^\perp
= \frac{\sigma}{\epsilon_0}.}
\]
Derivation: Tangential Component
Apply Stokes's theorem to a thin rectangular loop (length \(\ell\),
height \(\epsilon\to 0\)) straddling the surface. Since
\(\nabla\times\mathbf{E} = \mathbf{0}\):
\[
E_{\text{above}}^\| \ell - E_{\text{below}}^\| \ell = 0.
\]
\[
\boxed{E_{\text{above}}^\| = E_{\text{below}}^\|.}
\]
Gaussian pillbox and rectangular loop for boundary conditions.
Combined Vector Form and Potential Conditions
The two conditions combine into:
\[
\mathbf{E}_{\text{above}} - \mathbf{E}_{\text{below}}
= \frac{\sigma}{\epsilon_0}\hat{\mathbf{n}}.
\]
Since \(V\) is defined via a line integral of \(\mathbf{E}\), and the tangential
component is continuous, \(V\) itself cannot jump:
\[
\boxed{V_{\text{above}} = V_{\text{below}}.}
\]
However, its normal derivative is discontinuous:
\[
\boxed{\frac{\partial V_{\text{above}}}{\partial n}
- \frac{\partial V_{\text{below}}}{\partial n}
= -\frac{\sigma}{\epsilon_0},}
\]
where \(\partial/\partial n \equiv \nabla V\cdot\hat{\mathbf{n}}\).
Special case — Conductor surface:
Inside a conductor \(\mathbf{E} = \mathbf{0}\), so:
\[
E_{\text{outside}}^\perp = \frac{\sigma}{\epsilon_0}, \qquad
\mathbf{E}_{\text{outside}}^\| = \mathbf{0}.
\]
The field just outside a conductor is purely normal.
Concept Checks — Boundary Conditions
Concept check.
Two dielectrics (\(\varepsilon_1\), \(\varepsilon_2\)). Across interface, which is correct?
(a) \(D_\perp\) and \(E_\parallel\) continuous
(b) \(E_\parallel\) continuous, \(D_\perp\) discontinuous
(c) \(D_\perp\) continuous, \(E_\parallel\) discontinuous
(d) Both discontinuous
Show answer
(a). \(D_\perp\) is continuous (no free charges); \(E_\parallel\) is continuous. Hence \(D_\perp\) and \(E_\parallel\) are both continuous (option a).
Concept check.
Medium 1 (\(\varepsilon_{r1}=2\), \(z>0\)) and Medium 2 (\(\varepsilon_{r2}=3\), \(z<0\)). Field in medium 1: \(\vec{E}_1=2\hat{x}+3\hat{y}+4\hat{z}\). Interface charge-free. \(\vec{D}_2\) in medium 2:
(a) \(3\eps(2\hat{x}+3\hat{y}+\frac{8}{3}\hat{z})\)
(b) \(\eps(6\hat{x}+9\hat{y}+8\hat{z})\)
(c) \(\eps(4\hat{x}+6\hat{y}+8\hat{z})\)
(d) \(3\eps(2\hat{x}+3\hat{y}+4\hat{z})\)
Spherical dielectric shell (\(R_1<r<R_2\)), polarization \(\vec{P}=\dfrac{k}{r}\hat{r}\). Which is correct?
(a) \(\sigma_b(R_1)=-k/R_1\), \(\sigma_b(R_2)=k/R_2\), \(\rho_b=0\)
(b) \(\sigma_b=0\) both surfaces, \(\rho_b=-k/r^2\)
(c) \(\sigma_b(R_1)=-k/R_1\), \(\sigma_b(R_2)=k/R_2\), \(\rho_b=-k/r^2\)
(d) \(\sigma_b=0\), \(\rho_b=k/r^2\)
Show answer
(c). \(\rho_b=-\nabla\cdot\vec{P}=-\dfrac{1}{r^2}\dfrac{d}{dr}\!\left(r^2\dfrac{k}{r}\right)=-\dfrac{k}{r^2}\); \(\sigma_b=\vec{P}\cdot\hat{n}\) gives \(+k/R_2\) on the outer surface and \(-k/R_1\) on the inner surface.
Energy of a Continuous Charge Distribution
\[
W = \frac{1}{2}\int\rho\,V\,d\tau
= \frac{\epsilon_0}{2}\left(\int_{\mathcal{V}} E^2\,d\tau
+ \oint_S V\mathbf{E}\cdot d\mathbf{a}\right).
\]
The surface term vanishes when integrated over all space, giving the
compact result:
\[
\boxed{W = \frac{\epsilon_0}{2}\int E^2\,d\tau.}
\]
This has the interpretation of energy stored in the electric field itself,
with energy density \(u = \tfrac{\epsilon_0}{2}E^2\).
Conductors
Basic properties of a conductor in electrostatic equilibrium:
\(\mathbf{E} = \mathbf{0}\) inside. (Any internal field would drive
free charges until they cancel it.)
\(\rho = 0\) inside. (Follows from \(\nabla\cdot\mathbf{E} = 0\) inside.)
Any net charge resides on the surface.
\(\mathbf{E}\) is perpendicular to the surface just outside.
Example 1.36. Find the energy of a uniformly charged spherical shell
(total charge \(q\), radius \(R\)).
Example 1.37. Two infinite conducting plates are separated by \(l\).
A point charge \(q\) is at distance \(x\) from plate 1. Find the induced
charge on each plate.
Solution: Replacing the point charge by a uniform sheet (justified
by the uniqueness theorem—total induced charges depend only on geometry):
\[
q_1 = -\frac{q}{l}(l-x), \qquad q_2 = -\frac{q}{l}\,x.
\]
Example 1.38. Find the electric force per unit area on the surface of a
conductor with surface charge density \(\sigma\).
Solution: Near the surface, \(E = \sigma/\epsilon_0\). Split into the
self-field \(E_1 = \sigma/2\epsilon_0\) (which does not exert a net force on
the element itself) and the remaining field \(E_2 = \sigma/2\epsilon_0\):
\[
\frac{dF}{dS} = \sigma\cdot E_2 = \frac{\sigma^2}{2\epsilon_0}.
\]
Example 1.39. A metal ball of radius \(R = 1.5\,\mathrm{cm}\) carries
charge \(q = 10\,\mu\mathrm{C}\). Find the total force on one hemisphere.
Example 1.40. An uncharged conducting ball (radius \(R\)) in a uniform
external field develops induced surface charge
\(\sigma = \sigma_0\cos\theta\). Find the force on the induced charge of one
sign.
Solution:
\[
dF_z = \frac{\sigma^2}{2\epsilon_0}\cos\theta\,dS
= \frac{\pi\sigma_0^2 R^2}{\epsilon_0}\sin\theta\cos\theta\,d\theta.
\]
Integrating over the upper hemisphere (\(\cos\theta: 1\to 0\)):
\[
F = \frac{\pi\sigma_0^2 R^2}{4\epsilon_0}.
\]
Special Techniques
Poisson's and Laplace's Equations
From \(\mathbf{E} = -\nabla V\) and \(\nabla\cdot\mathbf{E} = \rho/\epsilon_0\):
\[
\boxed{\nabla^2 V = -\frac{\rho}{\epsilon_0}} \quad \text{(Poisson's equation)},
\qquad
\boxed{\nabla^2 V = 0} \quad \text{(Laplace's equation, when $\rho = 0$)}.
\]
Laplace's Equation in One Dimension
\(d^2V/dx^2 = 0 \Rightarrow V(x) = mx + b\) (a straight line). Two properties:
\(\nabla^2 V = 0\).
Average-value property:
\(V(\mathbf{r}) = \dfrac{1}{4\pi R^2}\oint_{\text{sphere}}V\,da\).
Special Methods of Solution
Method of images
Separation of variables
Multipole expansion
Boundary conditions and uniqueness:
First uniqueness theorem: The solution to Laplace's
equation in a volume \(\mathcal{V}\) is uniquely determined by the
values of \(V\) on the boundary surface \(S\).
Second uniqueness theorem: In a volume \(\mathcal{V}\)
bounded by conductors containing a specified \(\rho\), the electric
field is uniquely determined if the total charge on each conductor
is given.
Example 1.41. Find the average potential over a spherical surface of
radius \(R\) due to a point charge \(q\) located inside. Show that
\(V_{\text{ave}} = V_{\text{center}} + Q_{\text{enc}}/(4\pi\epsilon_0 R)\).
Solution: Using \(r^2 = z^2 + R^2 - 2zR\cos\theta\):
\[
V_{\text{ave}}
= \frac{1}{4\pi R^2}\frac{q}{4\pi\epsilon_0}
\int(z^2+R^2-2zR\cos\theta)^{-1/2}R^2\sin\theta\,d\theta\,d\phi.
\]
Since \(z < R\), \(\sqrt{z^2+R^2-2zR} = R - z\), and the integral evaluates to
\(\dfrac{q}{4\pi\epsilon_0 R}\). For multiple internal charges and external
charges contributing \(V_{\text{center}}\):
\[
V_{\text{ave}} = V_{\text{center}} + \frac{Q_{\text{enc}}}{4\pi\epsilon_0 R}.
\]
Method of Images
A point charge \(q\) above a grounded conducting plane.
Classic image problem: A point charge \(q\) lies a distance \(d\) above
an infinite grounded conducting plane (see figure). Find \(V\)
everywhere above the plane.
Solution:
Image configuration: \(+q\) above, \(-q\) below the plane.
Replace the conductor with an image charge \(-q\) at distance \(d\) below the
plane (see figure). The potential for \(z \geq 0\):
\[
V(s,\phi,z) = \frac{q}{4\pi\epsilon_0}
\left[\frac{1}{\sqrt{s^2+(z-d)^2}}
- \frac{1}{\sqrt{s^2+(z+d)^2}}\right].
\]
This automatically satisfies \(\nabla^2 V = -\rho/\epsilon_0\) (in SI units)
and \(V = 0\) on \(z = 0\). By the uniqueness theorem, this is the
solution.
Warning:
The image solution is valid only outside the conductor (\(z \geq 0\)).
The image charge does not physically exist; \(\mathbf{E} = \mathbf{0}\) inside
the conductor. The potential energy of the charge+conductor system is
not equal to that of the two-charge system (the field is zero inside
for the former).
The “mirror” analogy for the method of images.
Example 1.42. A charge \(q\) is at distance \(Z\) from the centre of a
grounded sphere of radius \(R\) (see figure).
(a) Find the potential everywhere.
(b) Find the induced surface charge density.
(c) Find the potential energy.
Point charge \(q\) outside a grounded sphere of radius \(R\).
Solution:
(a) Place image charge \(q' = -qR/Z\) at \(z' = R^2/Z\) inside the sphere
(see figure):
\[
z' = \frac{R^2}{Z}, \qquad q' = -\frac{qR}{Z}.
\]
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