$F = k\dfrac{q_1 q_2}{r^2}$, where $k = 9\times10^9$ N m² C⁻². Like charges repel, unlike attract. Newton's 3rd law holds — force on $q_1$ due to $q_2$ is equal and opposite to force on $q_2$ due to $q_1$.
Like charges — force is repulsive, equal and opposite by Newton's 3rd law
Q 1.1
Two small charged spheres: $q_1 = 2\times10^{-7}$ C, $q_2 = 3\times10^{-7}$ C, placed 30 cm apart in air. Find force.
Newton's 3rd Law: Force on B due to A = $0.2$ N, directed toward A (attractive). Equal in magnitude, opposite in direction.
Q 1.12
Two conducting spheres A and B, each with charge $q = 6.5\times10^{-7}$ C, centres 50 cm apart. (a) Find repulsive force. (b) What if each charge is doubled and distance halved?
Spheres A and B from Q 1.12 (identical size, $q = 6.5\times10^{-7}$ C each). An uncharged sphere C (same size) touches A first, then B, then removed. New force between A and B?
$q_A = q_B = 6.5\times10^{-7}$ C, $r = 0.50$ m, C uncharged initially
①
C touches A: charge shared equally.
$$q_A' = q_C' = \frac{6.5\times10^{-7}}{2} = 3.25\times10^{-7}\text{ C each}$$
②
C then touches B: total charge $= 3.25 + 6.5 = 9.75\ (\times10^{-7})$ C, shared equally.
$$q_B'' = q_C'' = \frac{9.75\times10^{-7}}{2} = 4.875\times10^{-7}\text{ C}$$
Charge is quantised ($q = ne$), conserved, and additive. Electron charge $e = 1.6\times10^{-19}$ C. Electric force is $\sim10^{39}$ times stronger than gravitational force between a proton and electron.
Q 1.3
Show that the ratio $\dfrac{ke^2}{Gm_em_p}$ is dimensionless and find its value. What does it signify?
$k=9\times10^9,\ e=1.6\times10^{-19}$ C, $G=6.67\times10^{-11}$, $m_e=9.11\times10^{-31}$ kg, $m_p=1.67\times10^{-27}$ kg
①
Dimensions: $[ke^2] = \text{N m}^2\text{C}^{-2}\cdot\text{C}^2 = \text{N m}^2$. $[Gm_em_p] = \text{N m}^2\text{kg}^{-2}\cdot\text{kg}^2 = \text{N m}^2$. Ratio is dimensionless ✓
★ Answer
$$\frac{ke^2}{Gm_em_p} \approx 2.27\times10^{39}$$
Electric force between e⁻ and p⁺ is $2.27\times10^{39}$ times stronger than gravitational force between them
Q 1.4
(a) What is quantisation of electric charge? (b) Why can we ignore quantisation for macroscopic charges?
Part (a)
Quantisation: Electric charge always exists as integer multiples of a minimum charge $e = 1.6\times10^{-19}$ C. So $q = \pm ne$ where $n = 1, 2, 3, \ldots$
$$q = ne, \quad n \in \mathbb{Z}, \quad e = 1.6\times10^{-19}\text{ C}$$
Part (b)
For macroscopic charges (say $1\ \mu$C $= 10^{-6}$ C), the number of electrons involved is $n \sim 10^{13}$ — so the charge appears practically continuous. The "graininess" due to $e$ is completely negligible at that scale.
Q 1.5
When glass is rubbed with silk, charges appear on both. How is this consistent with conservation of charge?
Before rubbing: both glass and silk are electrically neutral — total charge $= 0$.
On rubbing, electrons transfer from glass to silk (or vice versa). Glass gains $+Q$, silk gains $-Q$.
Total charge after $= +Q + (-Q) = 0$ — same as before. Charge is neither created nor destroyed, only transferred. This is perfect consistency with the Law of Conservation of Charge.
Q 1.11
A polythene piece rubbed with wool has charge $-3\times10^{-7}$ C. (a) Estimate number of electrons transferred. (b) Is there a transfer of mass?
$q = 3\times10^{-7}$ C (negative on polythene), $e = 1.6\times10^{-19}$ C, $m_e = 9.11\times10^{-31}$ kg
★ (a)
$$n \approx 1.875\times10^{12}\text{ electrons transferred from wool to polythene}$$
Part (b) — Mass transferred
①
$$\Delta m = n\cdot m_e = 1.875\times10^{12}\times9.11\times10^{-31} \approx 1.71\times10^{-18}\text{ kg}$$
Yes — a tiny mass is transferred (wool loses, polythene gains), but it is so negligibly small ($\sim10^{-18}$ kg) that it is undetectable in practice.
✦
03
Electric Field & Dipole
Q 1.6 · 1.7 · 1.8 · 1.9 · 1.10
$E = kq/r^2$ for a point charge. For a dipole: $p = q\cdot2a$. Torque on dipole: $\tau = pE\sin\theta$. Fields from two equal opposite charges at a midpoint add up (same direction).
Q 1.6
Four charges at corners of square ABCD (side 10 cm): $q_A = 2\ \mu$C, $q_B = -5\ \mu$C, $q_C = 2\ \mu$C, $q_D = -5\ \mu$C. Force on $1\ \mu$C charge at centre?
Fig 3.1 — A and C are diagonally opposite (+2 μC each); B and D are diagonally opposite (−5 μC each)
①
A and C are at opposite corners — equal charges, equal distances to O, forces on the 1 μC charge are equal and exactly opposite. They cancel.
②
B and D are at opposite corners — equal charges (−5 μC each), equal distances to O, forces are equal and exactly opposite. They cancel.
(a) Why can't a field line have sudden breaks? (b) Why can't two field lines cross?
Part (a)
A field line represents the path along which a positive test charge would move. A sudden break would imply the charge disappears and reappears — physically impossible. Also, at a break, the field would be undefined. Field lines must be continuous.
Part (b)
If two field lines crossed at a point P, then at P there would be two different directions of the electric field simultaneously. But the electric field at any point has a unique direction. Contradiction — hence field lines never cross.
Q 1.8
$q_A = 3\ \mu$C and $q_B = -3\ \mu$C are 20 cm apart in vacuum. (a) Electric field at midpoint O? (b) Force on $-1.5\times10^{-9}$ C test charge at O?
Fig 3.2 — Both E_A and E_B point in the same direction (A to B) at midpoint O
$|q_A| = |q_B| = 3\times10^{-6}$ C, $r = 0.10$ m (each to O)
$\phi = EA\cos\theta$. Gauss's Law: $\phi = q_{\text{enc}}/\varepsilon_0$. The net flux through a closed surface depends only on the enclosed charge — not on the shape or size of the surface.
Q 1.14
Three charged particles move in a uniform electrostatic field (plates). Give signs of charges and identify which has highest charge-to-mass ratio.
Fig 4.1 — Particles enter from left; field E points from +ve plate (top) to −ve plate (bottom)
Particle 3 has the highest $q/m$ ratio (maximum curvature → maximum deflection for same field).
Q 1.15
Uniform field $\vec{E} = 3\times10^3\,\hat{i}$ N/C. Square of side 10 cm. (a) Flux when plane is parallel to $yz$-plane? (b) Flux when normal makes $60°$ with $x$-axis?
$E = 3\times10^3$ N/C, side $= 0.10$ m, $A = 0.01$ m²
Part (a) — Normal along x (parallel to yz-plane)
①
Normal to the square is along $\hat{i}$; angle between E and normal $= 0°$.
$$\phi = EA\cos 0° = 3\times10^3\times0.01\times1$$
No — zero net flux means zero net charge inside. There could be equal amounts of positive and negative charges inside, which give net zero. We cannot conclude the box is empty.
Q 1.18
A point charge $+10\ \mu$C is 5 cm directly above the centre of a square of side 10 cm. Flux through the square?
Fig 4.2 — Trick: treat square as one face of a cube; charge is at centre of cube
①
Key trick: The charge is 5 cm above the centre of a 10 cm square — it's at the centre of an imaginary cube of edge 10 cm. The square is one face of this cube.
②
Total flux from the cube (by Gauss's Law):
$$\phi_{\text{total}} = \frac{q}{\varepsilon_0} = \frac{10\times10^{-6}}{8.854\times10^{-12}} = 1.13\times10^6\text{ N m}^2\text{C}^{-1}$$
③
By symmetry, flux is distributed equally among all 6 faces:
$$\phi_{\text{square}} = \frac{\phi_{\text{total}}}{6}$$
★ Answer
$$\phi_{\text{square}} = \frac{1.13\times10^6}{6} \approx 1.88\times10^5\text{ N m}^2\text{C}^{-1}$$
Q 1.19
A point charge $2.0\ \mu$C at centre of a cubic Gaussian surface (edge 9.0 cm). Net electric flux?
①
By Gauss's Law, flux depends only on enclosed charge, not on size or shape of surface:
$$\phi = \frac{q_{\text{enc}}}{\varepsilon_0} = \frac{2.0\times10^{-6}}{8.854\times10^{-12}}$$
★ Answer
$$\phi = 2.26\times10^5\text{ N m}^2\text{C}^{-1}$$
Q 1.20
Flux through a spherical Gaussian surface of radius 10 cm is $-1.0\times10^3$ N m² C⁻¹. (a) Flux if radius doubled? (b) Value of point charge?
Part (a)
Gauss's Law: $\phi = q/\varepsilon_0$ — depends only on charge, not on radius. Doubling the radius changes nothing.
★ (a)
$$\phi \text{ remains } = -1.0\times10^3\text{ N m}^2\text{C}^{-1}$$
Gauss's Law gives us $E$ for highly symmetric charge distributions. For a conducting sphere: $E = kq/r^2$ outside, $E = 0$ inside. For infinite line charge: $E = \lambda/2\pi\varepsilon_0 r$. For infinite plane: $E = \sigma/2\varepsilon_0$ (each side).
Q 1.21
Conducting sphere of radius 10 cm. Electric field 20 cm from centre is $1.5\times10^3$ N/C, pointing radially inward. Find net charge on sphere.
Uniformly charged conducting sphere of diameter 2.4 m, surface charge density $\sigma = 80.0\ \mu$C/m². (a) Charge on sphere. (b) Total electric flux leaving surface.
Two large thin parallel plates with surface charge densities $+\sigma$ and $-\sigma$ on their inner faces ($\sigma = 17.0\times10^{-22}$ C/m²). Find E in: (a) outer region of plate 1, (b) outer region of plate 2, (c) between the plates.
Fig 5.1 — Fields from each plate cancel outside, add up between the plates
①
Each infinite sheet produces $E = \sigma/2\varepsilon_0$ on each side. In the outer regions, the two fields (one from each plate) are equal and opposite — they cancel.
$$E_{\text{outer}} = \frac{\sigma}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} = 0$$
②
Between the plates, both fields point in the same direction (from $+$ to $-$):
$$E_{\text{between}} = \frac{\sigma}{2\varepsilon_0} + \frac{\sigma}{2\varepsilon_0} = \frac{\sigma}{\varepsilon_0}$$
$$= \frac{17\times10^{-22}}{8.854\times10^{-12}} \approx 1.92\times10^{-10}\text{ N/C}$$
★ Answer
(a) Outer region, Plate 1: $E = 0$
(b) Outer region, Plate 2: $E = 0$
(c) Between plates: $E = \sigma/\varepsilon_0 \approx 1.92\times10^{-10}$ N/C
