This page collects the Electric Charges and Fields Class 12 formulas from NCERT Physics Part I, Chapter 1 — Coulomb’s law, electric field, electric flux, the electric dipole, Gauss’s law, and the fields of the standard charge configurations obtained from them.
Each formula is grouped under the textbook sub-topic it belongs to, with the meaning and SI unit of every symbol, a when-to-use guide, three worked examples with original numbers, and the common mistakes that slip in while applying these formulas. For revision across the book, see the Class 12 physics formulas hub.
Formulas at a Glance
Every formula on this page in one compact index; the same formulas are grouped by sub-topic in the next section.
| Purpose (what you are finding) | Formula |
|---|---|
| Charge of a body in terms of electrons (quantisation) | \( q = ne \) |
| Total charge of a system (additivity) | \( q_{\text{total}} = q_1 + q_2 + \dots + q_n \) |
| Magnitude of the force between two point charges | \( F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_1 q_2|}{r^2} \) |
| Coulomb’s law as a vector | \( \mathbf{F}_{21} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{21}^2}\hat{\mathbf{r}}_{21} \) |
| Coulomb constant | \( k = \dfrac{1}{4\pi\varepsilon_0} \approx 9 \times 10^{9}\ \text{N m}^2\text{ C}^{-2} \) |
| Permittivity of free space | \( \varepsilon_0 = 8.854 \times 10^{-12}\ \text{C}^2\text{ N}^{-1}\text{ m}^{-2} \) |
| Electric force compared with gravity (electron–proton) | \( \dfrac{F_e}{F_G} = \dfrac{k e^2}{G m_e m_p} \approx 2.4 \times 10^{39} \) |
| Force on one charge due to many others (superposition) | \( \mathbf{F}_1 = \dfrac{q_1}{4\pi\varepsilon_0}\sum_{i=2}^{n}\dfrac{q_i}{r_{1i}^2}\hat{\mathbf{r}}_{1i} \) |
| Electric field of a point charge | \( \mathbf{E}(\mathbf{r}) = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}\hat{\mathbf{r}} \) |
| Force on a charge placed in an electric field | \( \mathbf{F} = q\mathbf{E} \) |
| Definition of electric field using a test charge | \( \mathbf{E} = \lim_{q \to 0}\dfrac{\mathbf{F}}{q} \) |
| Field of a system of charges | \( \mathbf{E}(\mathbf{r}) = \dfrac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n}\dfrac{q_i}{r_{iP}^2}\hat{\mathbf{r}}_{iP} \) |
| Vector area element (outward normal for a closed surface) | \( \Delta\mathbf{S} = \Delta S\, \hat{\mathbf{n}} \) |
| Flux through a small area element | \( \Delta\phi = \mathbf{E}\cdot\Delta\mathbf{S} = E\, \Delta S \cos\theta \) |
| Total flux through a surface | \( \phi = \sum \mathbf{E}\cdot\Delta\mathbf{S} \) |
| Dipole moment | \( \mathbf{p} = q(2a)\,\hat{\mathbf{p}}, \quad p = 2qa \) |
| Dipole field on the axis — exact | \( \mathbf{E} = \dfrac{q}{4\pi\varepsilon_0}\dfrac{4ar}{(r^2-a^2)^2}\hat{\mathbf{p}} \) |
| Dipole field on the axis — far field (\( r \gg a \)) | \( \mathbf{E} = \dfrac{2\mathbf{p}}{4\pi\varepsilon_0 r^3} \) |
| Dipole field on the equatorial plane — exact | \( \mathbf{E} = -\dfrac{2qa}{4\pi\varepsilon_0(r^2+a^2)^{3/2}}\hat{\mathbf{p}} \) |
| Dipole field on the equatorial plane — far field (\( r \gg a \)) | \( \mathbf{E} = -\dfrac{\mathbf{p}}{4\pi\varepsilon_0 r^3} \) |
| Torque on a dipole in a uniform field | \( \boldsymbol{\tau} = \mathbf{p}\times\mathbf{E}, \quad \tau = pE\sin\theta \) |
| Linear, surface and volume charge densities | \( \lambda = \dfrac{\Delta Q}{\Delta l}, \quad \sigma = \dfrac{\Delta Q}{\Delta S}, \quad \rho = \dfrac{\Delta Q}{\Delta V} \) |
| Field of a continuous volume distribution | \( \Delta\mathbf{E} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{\rho\, \Delta V}{r’^2}\hat{\mathbf{r}}’ \) |
| Gauss’s law (closed surface, \( q \) = charge enclosed) | \( \phi = \dfrac{q}{\varepsilon_0} \) |
| Field of an infinitely long charged wire | \( \mathbf{E} = \dfrac{\lambda}{2\pi\varepsilon_0 r}\hat{\mathbf{n}} \) |
| Field of an infinite charged plane sheet | \( \mathbf{E} = \dfrac{\sigma}{2\varepsilon_0}\hat{\mathbf{n}} \) |
| Field outside and inside a charged spherical shell | \( \mathbf{E} = \dfrac{q}{4\pi\varepsilon_0 r^2}\hat{\mathbf{r}}\ (r \geq R); \quad \mathbf{E} = 0\ (r \lt R) \) |
All Formulas, Grouped by Topic
Quantisation and Additivity of Charge
Charge on any body is an integral multiple of the elementary charge (NCERT, p. 5):
\[ q = ne \]
with \( n = 0, \pm 1, \pm 2, \dots \) and \( e = 1.602192 \times 10^{-19}\ \text{C} \approx 1.6 \times 10^{-19}\ \text{C} \). Charge is added algebraically, signs included (NCERT, p. 4):
\[ q_{\text{total}} = q_1 + q_2 + q_3 + \dots + q_n \]
The net charge of an isolated system is conserved; rubbing only transfers electrons from one body to another.
Coulomb’s Law
Force between two point charges in vacuum (NCERT, p. 8):
\[ F = \frac{1}{4\pi\varepsilon_0}\frac{|q_1 q_2|}{r^2} \]
Vector form — force on \( q_2 \) due to \( q_1 \) (NCERT, p. 8):
\[ \mathbf{F}_{21} = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_{21}^2}\hat{\mathbf{r}}_{21} \]
Similarly \( \mathbf{F}_{12} = -\mathbf{F}_{21} \), so Coulomb’s law agrees with Newton’s third law. Constants:
\[ k = \frac{1}{4\pi\varepsilon_0} \approx 9 \times 10^{9}\ \text{N m}^2\text{ C}^{-2}, \qquad \varepsilon_0 = 8.854 \times 10^{-12}\ \text{C}^2\text{ N}^{-1}\text{ m}^{-2} \]
Why \( 1/r^2 \)? The same set of field lines crosses every sphere centred on the charge, and the sphere’s area grows as \( r^2 \), so the field strength falls as \( 1/r^2 \) (NCERT, p. 20). Electric force is enormously stronger than gravity; for an electron–proton pair (NCERT, p. 38):
\[ \frac{F_e}{F_G} = \frac{k e^2}{G m_e m_p} \approx 2.4 \times 10^{39} \]
Forces Between Multiple Charges
Superposition principle — each pair acts independently, and the total force is the vector sum (NCERT, p. 12):
\[ \mathbf{F}_1 = \frac{q_1}{4\pi\varepsilon_0}\sum_{i=2}^{n}\frac{q_i}{r_{1i}^2}\hat{\mathbf{r}}_{1i} \]
Electric Field
Field of a point charge \( Q \) (NCERT, p. 14):
\[ \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r^2}\hat{\mathbf{r}} \]
Force on a charge in a field, and the operational definition of field with a vanishingly small test charge (NCERT, p. 14):
\[ \mathbf{F} = q\mathbf{E}, \qquad \mathbf{E} = \lim_{q \to 0}\frac{\mathbf{F}}{q} \]
Field of a system of charges — vector sum of the individual fields (NCERT, p. 15):
\[ \mathbf{E}(\mathbf{r}) = \frac{1}{4\pi\varepsilon_0}\sum_{i=1}^{n}\frac{q_i}{r_{iP}^2}\hat{\mathbf{r}}_{iP} \]
Electric Flux
Flux through a small planar element (NCERT, p. 22), with \( \theta \) the angle between \( \mathbf{E} \) and the area normal:
\[ \Delta\phi = \mathbf{E}\cdot\Delta\mathbf{S} = E\, \Delta S \cos\theta \]
For a curved surface, divide it into small elements and add; for a closed surface the normal is the outward normal (NCERT, p. 22):
\[ \phi = \sum \mathbf{E}\cdot\Delta\mathbf{S} \]
Electric Dipole
A dipole is a pair \( +q, -q \) separated by \( 2a \). Dipole moment, directed from \( -q \) to \( +q \) (NCERT, p. 24):
\[ \mathbf{p} = q(2a)\,\hat{\mathbf{p}}, \qquad p = 2qa \]
Field on the axis (NCERT, pp. 23–24):
\[ \mathbf{E} = \frac{q}{4\pi\varepsilon_0}\frac{4ar}{(r^2-a^2)^2}\hat{\mathbf{p}}, \qquad \mathbf{E} = \frac{2\mathbf{p}}{4\pi\varepsilon_0 r^3}\ \ (r \gg a) \]
Field on the equatorial plane (NCERT, pp. 24–25):
\[ \mathbf{E} = -\frac{2qa}{4\pi\varepsilon_0(r^2+a^2)^{3/2}}\hat{\mathbf{p}}, \qquad \mathbf{E} = -\frac{\mathbf{p}}{4\pi\varepsilon_0 r^3}\ \ (r \gg a) \]
At large distances the dipole field depends only on the product \( q(2a) = p \), not on \( q \) and \( a \) separately, and it falls as \( 1/r^3 \) — faster than a point charge’s \( 1/r^2 \).
Dipole in a Uniform External Field
The forces \( q\mathbf{E} \) and \( -q\mathbf{E} \) form a couple: net force is zero, but there is a torque that aligns the dipole with the field (NCERT, p. 27):
\[ \boldsymbol{\tau} = \mathbf{p}\times\mathbf{E}, \qquad \tau = pE\sin\theta = 2qaE\sin\theta \]
Continuous Charge Distribution
Charge densities for line, surface and volume elements (NCERT, p. 28):
\[ \lambda = \frac{\Delta Q}{\Delta l}, \qquad \sigma = \frac{\Delta Q}{\Delta S}, \qquad \rho = \frac{\Delta Q}{\Delta V} \]
Field of a small volume element, summed over the whole distribution (NCERT, p. 28):
\[ \Delta\mathbf{E} = \frac{1}{4\pi\varepsilon_0}\frac{\rho\, \Delta V}{r’^2}\hat{\mathbf{r}}’ \]
Gauss’s Law
Total electric flux through any closed surface \( S \) equals the net charge enclosed divided by \( \varepsilon_0 \) (NCERT, p. 30):
\[ \phi = \frac{q}{\varepsilon_0} \]
The field on the left is due to all charges; only the enclosed charge appears on the right. Gauss’s law is a consequence of the inverse-square form of Coulomb’s law — a violation of Gauss’s law would signal a departure from that law (NCERT, p. 30).
Applications of Gauss’s Law
Infinitely long straight wire of linear charge density \( \lambda \) (NCERT, p. 34):
\[ \mathbf{E} = \frac{\lambda}{2\pi\varepsilon_0 r}\hat{\mathbf{n}} \]
Infinite plane sheet of surface charge density \( \sigma \) (NCERT, p. 34) — the field is uniform on each side and independent of distance:
\[ \mathbf{E} = \frac{\sigma}{2\varepsilon_0}\hat{\mathbf{n}} \]
Thin uniformly charged spherical shell of radius \( R \) (NCERT, p. 35):
\[ \mathbf{E} = \frac{q}{4\pi\varepsilon_0 r^2}\hat{\mathbf{r}}\ \ (r \geq R), \qquad \mathbf{E} = 0\ \ (r \lt R) \]
Outside the shell the field behaves as if the total charge were concentrated at the centre; inside it is zero — an experimental check of the \( 1/r^2 \) law (NCERT, p. 36).
What Each Symbol Means
| Symbol | What it means | SI unit | Dimensions (NCERT, p. 40) |
|---|---|---|---|
| \( q, q_1, q_2, Q \) | point charge; \( Q \) usually the source charge | \( \text{C} \) | — |
| \( e \) | elementary charge (charge of a proton is \( +e \), of an electron \( -e \)) | \( \text{C} \) | — |
| \( n \) | integer \( 0, \pm 1, \pm 2, \dots \); number of electrons transferred | dimensionless (a count) | — |
| \( r \) | distance between charges, or distance from the source | \( \text{m} \) | — |
| \( r_{21} \) | distance from charge 1 to charge 2 | \( \text{m} \) | — |
| \( \hat{\mathbf{r}}_{21} \) | unit vector pointing from \( q_1 \) to \( q_2 \) | dimensionless | — |
| \( F, \mathbf{F}_{21} \) | electrostatic force (magnitude / vector on \( q_2 \) due to \( q_1 \)) | \( \text{N} \) | — |
| \( k \) | Coulomb constant \( = 1/4\pi\varepsilon_0 \) | \( \text{N m}^2\text{ C}^{-2} \) | — |
| \( \varepsilon_0 \) | permittivity of free space | \( \text{C}^2\text{ N}^{-1}\text{ m}^{-2} \) | — |
| \( \mathbf{E} \) | electric field | \( \text{N C}^{-1} \) (also written \( \text{V m}^{-1} \)) | \( [\text{MLT}^{-3}\text{A}^{-1}] \) |
| \( \mathbf{p}, p \) | electric dipole moment, from \( -q \) to \( +q \) | \( \text{C m} \) | \( [\text{LTA}] \) |
| \( 2a \) | separation between the two charges of a dipole | \( \text{m} \) | — |
| \( \boldsymbol{\tau}, \tau \) | torque on a dipole in an electric field | \( \text{N m} \) | — |
| \( \theta \) | angle between \( \mathbf{p} \) and \( \mathbf{E} \), or between \( \mathbf{E} \) and the area normal | dimensionless (angle) | — |
| \( \Delta\mathbf{S}, \Delta S \) | area element (vector / magnitude) | \( \text{m}^2 \) | \( [\text{L}^2] \) |
| \( \phi, \Delta\phi \) | electric flux | \( \text{N m}^2\text{ C}^{-1} \) (also \( \text{V m} \)) | \( [\text{ML}^3\text{T}^{-3}\text{A}^{-1}] \) |
| \( \lambda \) | linear charge density | \( \text{C m}^{-1} \) | \( [\text{L}^{-1}\text{TA}] \) |
| \( \sigma \) | surface charge density | \( \text{C m}^{-2} \) | \( [\text{L}^{-2}\text{TA}] \) |
| \( \rho \) | volume charge density | \( \text{C m}^{-3} \) | \( [\text{L}^{-3}\text{TA}] \) |
| \( \Delta Q \) | charge contained in a small element | \( \text{C} \) | — |
| \( \hat{\mathbf{n}} \) | unit vector normal to a surface or along the radial direction | dimensionless | — |
| \( G \) | gravitational constant (used in the force-ratio formula) | \( \text{N m}^2\text{ kg}^{-2} \) | — |
When to Use Each Formula
| Formula | Use it when | Condition to check |
|---|---|---|
| \( q = ne \) | charge is asked in terms of electrons gained or lost | \( n \) is an integer; sign tells excess (+) or deficit (−) |
| \( F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{|q_1 q_2|}{r^2} \) | force between two point charges, distance given | charges are point-like (size \( \ll r \)); vacuum |
| \( \mathbf{F}_{21} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q_1 q_2}{r_{21}^2}\hat{\mathbf{r}}_{21} \) | force needed as a vector, direction along the join | \( q_1 q_2 \gt 0 \) repulsion; \( q_1 q_2 \lt 0 \) attraction |
| \( \mathbf{F}_1 = \dfrac{q_1}{4\pi\varepsilon_0}\sum \dfrac{q_i}{r_{1i}^2}\hat{\mathbf{r}}_{1i} \) | force on one charge due to several others | add pair-wise Coulomb forces as vectors |
| \( \mathbf{E} = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}\hat{\mathbf{r}} \) | field at distance \( r \) from a single point charge | radially outward for \( Q \gt 0 \), inward for \( Q \lt 0 \) |
| \( \mathbf{F} = q\mathbf{E} \) | force on a charge placed in a known field | use \( q \) with its sign; \( E \) in \( \text{N C}^{-1} \) |
| \( \Delta\phi = E\, \Delta S \cos\theta \) | flux through one small plane area | \( \theta \) between \( \mathbf{E} \) and the area normal |
| \( \phi = q/\varepsilon_0 \) | total flux through a closed surface, or the charge enclosed | Gaussian surface must not pass through discrete charges |
| \( \mathbf{p} = q(2a)\,\hat{\mathbf{p}} \) | dipole moment from charges and separation | direction from \( -q \) to \( +q \) |
| \( E = \dfrac{2p}{4\pi\varepsilon_0 r^3} \) | field on the dipole axis, far from the dipole | \( r \gg a \); along \( \mathbf{p} \) |
| \( E = \dfrac{p}{4\pi\varepsilon_0 r^3} \) | field on the perpendicular bisector of the dipole | \( r \gg a \); opposite to \( \mathbf{p} \) |
| \( \tau = pE\sin\theta \) | torque on a dipole in a uniform external field | net force is zero; torque aligns \( \mathbf{p} \) with \( \mathbf{E} \) |
| \( \lambda, \sigma, \rho \) | convert element charge to density; field of continuous distributions | element small macroscopically, large microscopically |
| \( E = \dfrac{\lambda}{2\pi\varepsilon_0 r} \) | field near a long straight charged wire | infinite wire; \( r \) is perpendicular distance |
| \( E = \dfrac{\sigma}{2\varepsilon_0} \) | field near a large charged plane sheet | infinite sheet; uniform field on each side |
| \( E = \dfrac{q}{4\pi\varepsilon_0 r^2} \), \( E = 0 \) | field outside / inside a charged spherical shell | outside acts like a point charge at the centre; inside zero |
Worked Examples
Three typical applications of the electric charges and fields class 12 formulas, with original numbers. For more practice on the textbook’s own questions, revise through the physics formulas library.
Example 1: Force between two point charges (Coulomb’s law)
Step 1: A charge \( q_1 = +3.0\ \mu\text{C} \) and a charge \( q_2 = -4.0\ \mu\text{C} \) are 20 cm apart in vacuum.
Use the magnitude form of Coulomb’s law with \( 1/4\pi\varepsilon_0 = 9 \times 10^{9}\ \text{N m}^2\text{ C}^{-2} \).
\[ F = \frac{1}{4\pi\varepsilon_0}\frac{|q_1 q_2|}{r^2} \]
Step 2: Convert to SI: \( q_1 = 3.0 \times 10^{-6}\ \text{C} \), \( |q_2| = 4.0 \times 10^{-6}\ \text{C} \), \( r = 0.20\ \text{m} \).
\[ F = \frac{(9 \times 10^{9})(3.0 \times 10^{-6})(4.0 \times 10^{-6})}{(0.20)^2} = \frac{0.108}{0.040} = 2.7\ \text{N} \]
Step 3: Signs: the charges are unlike, so the force is attractive.
Final answer: each charge experiences \( 2.7\ \text{N} \), attractive, along the line joining them.
Example 2: Finding enclosed charge from flux (Gauss’s law)
Step 1: The net outward electric flux through a closed surface is \( 2.26 \times 10^{3}\ \text{N m}^2\text{ C}^{-1} \).
Use Gauss’s law and solve for \( q \).
\[ \phi = \frac{q}{\varepsilon_0} \quad \Rightarrow \quad q = \phi\, \varepsilon_0 \]
Step 2: Substitute \( \varepsilon_0 = 8.854 \times 10^{-12}\ \text{C}^2\text{ N}^{-1}\text{ m}^{-2} \).
\[ q = (2.26 \times 10^{3})(8.854 \times 10^{-12}) = 2.0 \times 10^{-8}\ \text{C} \]
Step 3: The flux is outward, so the enclosed charge is positive.
Final answer: net charge inside the surface is \( +2.0 \times 10^{-8}\ \text{C} \).
Example 3: Dipole field far from the dipole
Step 1: A dipole has charges \( \pm 2.0\ \mu\text{C} \) separated by \( 2a = 4.0\ \text{mm} \).
Find \( \mathbf{E} \) at \( r = 12\ \text{cm} \) from the centre on the axis, on the side of the positive charge.
First get the dipole moment.
\[ p = q(2a) = (2.0 \times 10^{-6})(4.0 \times 10^{-3}) = 8.0 \times 10^{-9}\ \text{C m} \]
Step 2: Check the far-field condition: \( r/a = 0.12/0.002 = 60 \gg 1 \), so use the axial far-field form.
\[ E = \frac{2p}{4\pi\varepsilon_0 r^3} = \frac{2(9 \times 10^{9})(8.0 \times 10^{-9})}{(0.12)^3} = \frac{144}{1.73 \times 10^{-3}} = 8.3 \times 10^{4}\ \text{N C}^{-1} \]
Step 3: On the axis the field points along \( \mathbf{p} \), i.e.
from \( -q \) toward \( +q \).
Final answer: \( 8.3 \times 10^{4}\ \text{N C}^{-1} \) along the dipole axis from the negative to the positive charge.
Common Mistakes to Avoid
These are the errors that appear again and again when the formulas of this chapter are applied in numericals.
| Mistake | Correct rule | How to check your answer |
|---|---|---|
| Putting signed charges into the magnitude formula and getting a negative \( F \). | Use \( |q_1 q_2| \) in \( F \); the sign of \( q_1 q_2 \) decides only the direction (like → repel, unlike → attract). | Your \( F \) is a positive number; state the direction in words. |
| Forgetting SI conversions — \( r \) in cm, \( q \) in \( \mu\text{C} \). | \( r \) in metres, \( q \) in coulombs: \( 1\ \mu\text{C} = 10^{-6}\ \text{C} \). | Force must come out in N, field in \( \text{N C}^{-1} \), flux in \( \text{N m}^2\text{ C}^{-1} \). |
| Counting outside charges in Gauss’s law. | Only the charge enclosed by the closed surface goes into \( q \); the field itself is due to all charges. | Zero flux means zero net enclosed charge, but \( \mathbf{E} \) may still be non-zero. |
| Treating the dipole far field as \( 1/r^2 \), or dropping the factor 2 on the axis. | Axis: \( \dfrac{2p}{4\pi\varepsilon_0 r^3} \); equatorial: \( -\dfrac{p}{4\pi\varepsilon_0 r^3} \). | Doubling \( r \) must reduce a dipole field by a factor of 8. |
| Mixing up \( \lambda, \sigma, \rho \), or using the wire/sheet/shell formula for the wrong geometry. | Wire → \( \lambda \) with \( 2\pi\varepsilon_0 r \); sheet → \( \sigma \) with \( 2\varepsilon_0 \); shell → \( q/4\pi\varepsilon_0 r^2 \) outside. | Dimension check: each combination gives \( \text{N C}^{-1} \). |
| Taking the dipole moment direction from \( +q \) to \( -q \). | \( \mathbf{p} \) points from \( -q \) to \( +q \). | On the axis, \( \mathbf{E} \) is along \( \mathbf{p} \); on the equator, opposite to \( \mathbf{p} \). |
Frequently Asked Questions
When can I apply Coulomb’s law as given in this chapter?
When the charges are point charges (size much smaller than their separation) and they are in vacuum. If the space between the charges is filled with matter, the force changes; that case is taken up in Chapter 2. State “in vacuum” when you quote the law.
Why does the dipole field fall off faster than a point-charge field?
Because the fields of \( +q \) and \( -q \) nearly cancel at large distances. The leading surviving term is proportional to the product \( q \times 2a = p \), which gives the \( 1/r^3 \) dependence instead of \( 1/r^2 \). A dipole’s field is therefore weaker than a single charge’s field at the same distance.
If the flux through a closed surface is zero, is the electric field zero on the surface?
No. Zero flux means the net charge enclosed is zero (Gauss’s law). Field lines may still enter and leave the surface, because the field is produced by all charges, inside and outside.
Why is the field of an infinite plane sheet independent of distance?
A Gaussian pill-box has two faces crossed by flux, so \( 2EA = \sigma A/\varepsilon_0 \), giving \( E = \sigma/2\varepsilon_0 \). Because the sheet is treated as infinite, the charge configuration looks identical at every distance, so the field is uniform on each side.
Reference: NCERT Class 12 Physics textbook (Rationalised NCERT), Chapter 1, Electric Charges and Fields. Verify the constants directly in the official NCERT textbook at ncert.nic.in.
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- Electric Charges and Fields Notes
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