Is Electric Charges and Fields part of the EAPCET Physics syllabus?

Yes. Electric Charges and Fields is taught in Intermediate second year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.

How many Physics questions are there in EAPCET?

40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.

What mistakes do students make in Electric Charges and Fields?

Adding the magnitudes of fields from several charges — add them as vectors, taking directions into account. Including charges outside a closed surface in Gauss’s law — only the enclosed charge decides the net flux. Forgetting the square root when locating a null point — the field ratio involves r², so distances scale as √(q₁/q₂). Saying a dipole in a uniform field has a net force — the forces on its two charges cancel, leaving only a torque.

EAPCET Physics · Intermediate second year

Electric Charges and Fields

Electric Charges and Fields covers Coulomb’s law, electric fields of charges and distributions, dipoles and Gauss’s law.

Subject: Physics
Taught in: Intermediate second year
Physics in the paper: 40 of 160 questions

Key formulas and results

$F = \frac{kq_{1}q_{2}}{r^{2}}$ , with $k = \frac{1}{4\pi\varepsilon_{0}} \approx 9 \times 10^{9}\ \text{N m}^{2}\,\text{C}^{-2}$
Field of a point charge: $E = \frac{kq}{r^{2}}$
Gauss’s law: $\Phi = \frac{q_{\text{enclosed}}}{\varepsilon_{0}}$
Infinite line charge: $E = \frac{\lambda}{2\pi\varepsilon_{0}r}$ ; infinite sheet: $E = \frac{\sigma}{2\varepsilon_{0}}$
Dipole moment $p = q \times 2a$ ; torque $\vec{\tau} = \vec{p} \times \vec{E}$
Charge is quantised: $q = ne$
Field on the axis of a uniformly charged ring (charge $Q$ , radius $R$ ) at distance $x$ : $E = \frac{kQx}{(R^{2} + x^{2})^{3/2}}$
Maximum field on the ring axis: at $x = R/\sqrt{2}$ , $E_{\max} = \frac{2kQ}{3\sqrt{3}R^{2}}$
Field on the axis of a uniformly charged disc: $E = \frac{\sigma}{2\varepsilon_{0}}\left[1 - \frac{x}{\sqrt{x^{2} + R^{2}}}\right] = 2\pi k\sigma(1 - \cos\theta)$ , $\theta$ = half-angle the disc subtends at the point
Field at the centre of a uniformly charged arc of angle $\phi$ : $E = \frac{2k\lambda}{R}\sin(\phi/2)$ ; semicircular ring: $E = \frac{2k\lambda}{R}$
Short dipole: axial field $E = \frac{2kp}{r^{3}}$ (along $\vec{p}$ ); equatorial field $E = \frac{kp}{r^{3}}$ (opposite to $\vec{p}$ )
Dipole field at a general point $(r, \theta)$ : $E = \frac{kp}{r^{3}}\sqrt{1 + 3\cos^{2}\theta}$ , making angle $\alpha$ with the radial line where $\tan\alpha = \tfrac{1}{2}\tan\theta$
Dipole in a uniform field: $U = -\vec{p}\cdot\vec{E} = -pE\cos\theta$ ; work to rotate from $\theta_{1}$ to $\theta_{2}$ : $W = pE(\cos\theta_{1} - \cos\theta_{2})$
Period of small oscillations of a dipole about stable equilibrium: $T = 2\pi\sqrt{I/pE}$

What this chapter covers

Properties of charge
Coulomb’s law
Electric field and field lines
Electric dipole
Electric flux and Gauss’s law
Fields of continuous distributions

How it is usually asked

Superposition of forces and fields
Flux through surfaces
Fields of charge distributions
Dipole in a field

Understanding Electric Charges and Fields

Electric charge is conserved and quantised in units of e. Coulomb’s law gives the force between two point charges, which acts along the line joining them and falls off as 1/r². The electric field at a point is the force per unit positive test charge; fields from several charges add as vectors (superposition), so directions matter as much as magnitudes.

Gauss’s law states that the total electric flux through any closed surface equals the enclosed charge divided by ε₀. Charges outside the surface contribute zero net flux. For symmetric distributions, such as a long line, a large sheet or a sphere, Gauss’s law gives the field quickly. An electric dipole in a uniform field feels a torque p × E but no net force.

EAPCET usually tests vector addition of forces or fields, null points, the flux through a cube or other closed surface, and the torque on a dipole. For a charge q at the centre of a cube, symmetry spreads the total flux q/ε₀ equally over the six faces, so each face receives q/(6ε₀); if the charge sits at a corner instead, the flux through the whole cube is q/(8ε₀).

Worked example

EAPCET-style question

Two point charges +2 μC and +8 μC are placed 0.3 m apart. At what point on the line joining them is the electric field zero?

(1)0.1 m from the 2 μC charge
(2)0.2 m from the 2 μC charge
(3)0.15 m from either charge
(4)0.06 m from the 2 μC charge

Solution

For two like charges, the null point lies between them, closer to the smaller charge.
Let it be $x$ from the 2 μC charge: $\frac{k(2)}{x^{2}} = \frac{k(8)}{(0.3 - x)^{2}}$ .
Taking square roots: $\frac{0.3 - x}{x} = \sqrt{8/2} = 2$ .
$0.3 - x = 2x$ , so $x = 0.1\ \text{m}$ from the 2 μC charge.

Answer: 0.1 m from the 2 μC charge

Shortcuts and results to remember

A charge $q$ at the centre of a cube sends flux $q/\varepsilon_{0}$ through the whole cube and $q/6\varepsilon_{0}$ through each face.
Net flux through a closed surface depends only on the charge enclosed; charges outside change E on the surface but not the total flux.
For an annular disc or a disc seen at an angle, use $E = 2\pi k\sigma(1 - \cos\theta)$ for each disc and subtract; a very large disc ( $\theta \to 90^\circ$ ) gives $\sigma/2\varepsilon_{0}$ .
A quarter-circle arc of charge gives $E = \frac{\sqrt{2}k\lambda}{R}$ at its centre, a semicircle $\frac{2k\lambda}{R}$ , and a full uniform ring zero.
Dipole in a uniform field: net force is always zero; $\theta = 0^\circ$ is stable ( $U = -pE$ , minimum), $\theta = 180^\circ$ is unstable ( $U = +pE$ ), $\theta = 90^\circ$ gives the maximum torque $pE$ .

Common mistakes to avoid

Adding the magnitudes of fields from several charges — add them as vectors, taking directions into account.
Including charges outside a closed surface in Gauss’s law — only the enclosed charge decides the net flux.
Forgetting the square root when locating a null point — the field ratio involves r², so distances scale as √(q₁/q₂).
Saying a dipole in a uniform field has a net force — the forces on its two charges cancel, leaving only a torque.

How to prepare it

Use Gauss’s law whenever the charge distribution has symmetry — it replaces integration.

Once the formulas feel familiar, test them the way the exam does: practise Electric Charges and Fields questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise Electric Charges and Fields chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Electric Charges and Fields part of the EAPCET Physics syllabus?: Yes. Electric Charges and Fields is taught in Intermediate second year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Electric Charges and Fields for EAPCET?: Use Gauss’s law whenever the charge distribution has symmetry — it replaces integration.
How many Physics questions are there in EAPCET?: 40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Electric Charges and Fields?: Adding the magnitudes of fields from several charges — add them as vectors, taking directions into account. Including charges outside a closed surface in Gauss’s law — only the enclosed charge decides the net flux. Forgetting the square root when locating a null point — the field ratio involves r², so distances scale as √(q₁/q₂). Saying a dipole in a uniform field has a net force — the forces on its two charges cancel, leaving only a torque.