Chapter 1: Electric Charges and Fields

The magnitude of charge carried by one electron is $$e = 1.6 \times 10^{-19} \, \mathrm{C}$$.

In one second, $$n = 10^9$$ electrons move out, so the charge transferred per second is

$$q_{1\,\mathrm{s}} = n e = 10^9 \times 1.6 \times 10^{-19} \, \mathrm{C} = 1.6 \times 10^{-10} \, \mathrm{C}$$

To accumulate a total charge $$Q = 1 \, \mathrm{C}$$, the time required is

$$t = \frac{Q}{q_{1\,\mathrm{s}}} = \frac{1}{1.6 \times 10^{-10}} \, \mathrm{s} = 6.25 \times 10^9 \, \mathrm{s}$$

Converting to years (one year $$= 365 \times 24 \times 3600 \approx 3.15 \times 10^7 \, \mathrm{s}$$):

$$t = \frac{6.25 \times 10^9}{3.15 \times 10^7} \, \mathrm{years} \approx 198 \, \mathrm{years}$$

Thus, even at the rate of $$10^9$$ electrons per second, it takes about 200 years to collect a charge of just one coulomb. This shows that one coulomb is a very large unit of charge.

Take the mass of one cup of water as $$250 \, \mathrm{g}$$. The molar mass of water is $$18 \, \mathrm{g}$$, and one mole contains $$N_A = 6.02 \times 10^{23}$$ molecules.

Number of water molecules in the cup:

$$N = \frac{250}{18} \times 6.02 \times 10^{23} \approx 8.36 \times 10^{24}$$

Each water molecule $$\mathrm{H_2O}$$ has two hydrogen atoms (1 proton each) and one oxygen atom (8 protons), i.e. $$10$$ protons and $$10$$ electrons. So each molecule carries a positive charge $$10e$$ and an equal negative charge $$10e$$.

Total positive charge:

$$Q_+ = N \times 10 \times e = \frac{250}{18} \times 6.02 \times 10^{23} \times 10 \times 1.6 \times 10^{-19} \, \mathrm{C}$$

$$Q_+ \approx 1.34 \times 10^7 \, \mathrm{C}$$

Since each molecule is electrically neutral, the total negative charge has the same magnitude:

$$Q_- \approx -1.34 \times 10^7 \, \mathrm{C}$$

The enormous magnitude shows that ordinary matter contains a huge amount of charge of both signs, almost exactly balanced.

(a) Ratio of electric force to gravitational force

Both forces obey an inverse-square law, so the distance $$r$$ cancels in the ratio.

(i) For an electron and a proton:

$$\left|\frac{F_e}{F_g}\right| = \frac{\dfrac{1}{4\pi\varepsilon_0}\dfrac{e^2}{r^2}}{\dfrac{G m_e m_p}{r^2}} = \frac{e^2}{4\pi\varepsilon_0\, G\, m_e m_p} = \frac{k e^2}{G\, m_e m_p}$$

Substituting $$k = 9 \times 10^9$$, $$e = 1.6 \times 10^{-19} \, \mathrm{C}$$, $$G = 6.67 \times 10^{-11}$$, $$m_e = 9.11 \times 10^{-31} \, \mathrm{kg}$$, $$m_p = 1.67 \times 10^{-27} \, \mathrm{kg}$$:

$$k e^2 = 9 \times 10^9 \times (1.6 \times 10^{-19})^2 = 2.3 \times 10^{-28}$$

$$G\, m_e m_p = 6.67 \times 10^{-11} \times 9.11 \times 10^{-31} \times 1.67 \times 10^{-27} \approx 1.0 \times 10^{-67}$$

$$\left|\frac{F_e}{F_g}\right| = \frac{2.3 \times 10^{-28}}{1.0 \times 10^{-67}} \approx 2.3 \times 10^{39}$$

(ii) For two protons: replace $$m_e$$ by $$m_p$$ in the denominator:

$$\left|\frac{F_e}{F_g}\right| = \frac{k e^2}{G\, m_p^2} = \frac{2.3 \times 10^{-28}}{6.67 \times 10^{-11} \times (1.67 \times 10^{-27})^2} \approx 1.3 \times 10^{36}$$

In both cases the electric force is enormously larger than the gravitational force.

(b) Accelerations at separation $$r = 10^{-10} \, \mathrm{m}$$

The magnitude of the electric (Coulomb) force is

$$F = \frac{k e^2}{r^2} = \frac{9 \times 10^9 \times (1.6 \times 10^{-19})^2}{(10^{-10})^2} = \frac{2.3 \times 10^{-28}}{10^{-20}} = 2.3 \times 10^{-8} \, \mathrm{N}$$

Acceleration of the electron:

$$a_e = \frac{F}{m_e} = \frac{2.3 \times 10^{-8}}{9.11 \times 10^{-31}} \approx 2.5 \times 10^{22} \, \mathrm{m\,s^{-2}}$$

Acceleration of the proton:

$$a_p = \frac{F}{m_p} = \frac{2.3 \times 10^{-8}}{1.67 \times 10^{-27}} \approx 1.4 \times 10^{19} \, \mathrm{m\,s^{-2}}$$

These accelerations are gigantic compared with $$g = 9.8 \, \mathrm{m\,s^{-2}}$$, confirming that gravity is utterly negligible for the motion of these particles under their mutual electric attraction.

Let the original charge on sphere A be $$q$$ and on sphere B be $$q'$$. With centres a distance $$r$$ apart, the Coulomb force on each (treating them as point charges) is

$$F = \frac{1}{4\pi\varepsilon_0}\frac{q\,q'}{r^2}$$

When the uncharged identical sphere C touches A, the charge redistributes equally between the two identical conductors. By symmetry each then carries $$q/2$$. Similarly, after the identical uncharged sphere D touches B, sphere B is left with $$q'/2$$.

Now C and D are removed and the separation is halved to $$r' = r/2$$. The new force is

$$F' = \frac{1}{4\pi\varepsilon_0}\frac{(q/2)(q'/2)}{(r/2)^2} = \frac{1}{4\pi\varepsilon_0}\frac{q q'/4}{r^2/4} = \frac{1}{4\pi\varepsilon_0}\frac{q q'}{r^2} = F$$

The factor $$1/4$$ from halving each charge is exactly cancelled by the factor $$1/4$$ from halving the distance. Hence the expected repulsion of A is the same as before, $$F' = F$$.

Let the side of the equilateral triangle be $$l$$, with charges $$q$$ at each vertex A, B, C and charge $$Q$$ at the centroid O.

Drop a perpendicular AD from A to side BC. Then $$AD = AC\cos 30^\circ = \dfrac{\sqrt3}{2}\,l$$. The centroid divides the median in the ratio $$2:1$$, so the distance of O from each vertex is

$$AO = BO = CO = \frac{2}{3}AD = \frac{2}{3}\cdot\frac{\sqrt3}{2}l = \frac{l}{\sqrt3}$$

The magnitude of the force on $$Q$$ due to a charge $$q$$ at any vertex is

$$F_0 = \frac{1}{4\pi\varepsilon_0}\frac{Qq}{(l/\sqrt3)^2} = \frac{3}{4\pi\varepsilon_0}\frac{Qq}{l^2}$$

So the three forces have equal magnitude $$\dfrac{3}{4\pi\varepsilon_0}\dfrac{Qq}{l^2}$$, directed along AO, BO and CO respectively (repulsive, since $$Q$$ and $$q$$ have the same sign).

Consider the forces $$\mathbf F_2$$ (along BO) and $$\mathbf F_3$$ (along CO). They are equal in magnitude and inclined at $$120^\circ$$; by the parallelogram law their resultant has magnitude $$\dfrac{3}{4\pi\varepsilon_0}\dfrac{Qq}{l^2}$$ and is directed along OA, i.e. exactly opposite to $$\mathbf F_1$$ (which is along AO).

Therefore the total force is

$$\mathbf F = \frac{3}{4\pi\varepsilon_0}\frac{Qq}{l^2}\,(\hat{\mathbf r} - \hat{\mathbf r}) = 0$$

where $$\hat{\mathbf r}$$ is the unit vector along OA. By the symmetry of the configuration the three equal forces, spaced $$120^\circ$$ apart, must add to zero.

Let $$F = \dfrac{1}{4\pi\varepsilon_0}\dfrac{q^2}{l^2}$$ be the magnitude of the Coulomb force between any pair of the charges (every pair is separated by the side length $$l$$ and the charge magnitudes are all $$q$$).

Force on charge $$q$$ at A: It is repelled by $$q$$ at B (force of magnitude $$F$$ along BA) and attracted by $$-q$$ at C (force of magnitude $$F$$ along AC). These two forces of equal magnitude $$F$$ are inclined at $$60^\circ$$ to each other, so their resultant has magnitude

$$F_1 = \sqrt{F^2 + F^2 + 2F^2\cos 60^\circ} = \sqrt{3F^2} = \sqrt{3}\,F$$

directed parallel to BC.

Force on charge $$q$$ at B: By identical reasoning (repelled by A, attracted by C), its magnitude is

$$F_2 = \sqrt{3}\,F$$

directed parallel to AC.

Force on charge $$-q$$ at C: It is attracted by $$q$$ at A and by $$q$$ at B. The two forces of magnitude $$F$$ are again $$60^\circ$$ apart, so

$$F_3 = \sqrt{3}\,F$$

directed along the bisector of angle BCA (pointing into the triangle).

Thus each charge experiences a force of magnitude

$$\sqrt3\,F = \frac{\sqrt3\,q^2}{4\pi\varepsilon_0 l^2}$$

The vector sum of the three forces is $$\mathbf F_1 + \mathbf F_2 + \mathbf F_3 = 0$$, consistent with Newton's third law applied to the internal forces of the system.

Electron [Fig. 1.10(a)]: The field points upward, so the negatively charged electron feels a downward force of magnitude $$eE$$. Its acceleration is

$$a_e = \frac{eE}{m_e}$$

Starting from rest, the time to fall a distance $$h$$ follows from $$h = \tfrac12 a_e t_e^2$$:

$$t_e = \sqrt{\frac{2h}{a_e}} = \sqrt{\frac{2h\,m_e}{eE}}$$

With $$h = 1.5 \times 10^{-2} \, \mathrm{m}$$, $$E = 2.0 \times 10^4 \, \mathrm{N\,C^{-1}}$$, $$e = 1.6 \times 10^{-19} \, \mathrm{C}$$ and $$m_e = 9.11 \times 10^{-31} \, \mathrm{kg}$$:

$$t_e = \sqrt{\frac{2 \times 1.5 \times 10^{-2} \times 9.11 \times 10^{-31}}{1.6 \times 10^{-19} \times 2.0 \times 10^4}} \approx 2.9 \times 10^{-9} \, \mathrm{s}$$

Proton [Fig. 1.10(b)]: The field is reversed (now downward), so the positively charged proton also feels a downward force $$eE$$, with acceleration

$$a_p = \frac{eE}{m_p}$$

The time of fall is

$$t_p = \sqrt{\frac{2h\,m_p}{eE}} = \sqrt{\frac{2 \times 1.5 \times 10^{-2} \times 1.67 \times 10^{-27}}{1.6 \times 10^{-19} \times 2.0 \times 10^4}} \approx 1.3 \times 10^{-7} \, \mathrm{s}$$

Contrast with free fall: Here the time of fall depends on the mass, $$t \propto \sqrt{m}$$, so the heavier proton takes far longer ($$t_p/t_e = \sqrt{m_p/m_e} \approx 43$$) to fall the same distance. In free fall under gravity the acceleration is $$g$$, independent of mass, so all bodies take the same time. (Also $$a_p = eE/m_p \approx 1.9 \times 10^{12} \, \mathrm{m\,s^{-2}}$$ is enormous compared with $$g$$, which justifies ignoring gravity in this problem.)

Take $$q_1 = +10^{-8} \, \mathrm{C}$$ on the left and $$q_2 = -10^{-8} \, \mathrm{C}$$ on the right, separated by $$0.1 \, \mathrm{m}$$. Use $$k = 9 \times 10^9 \, \mathrm{N\,m^2\,C^{-2}}$$.

At point A (the midpoint of the line, $$0.05 \, \mathrm{m}$$ from each charge):

The field $$\mathbf E_{1A}$$ of the positive charge $$q_1$$ points away from $$q_1$$ (towards the right); the field $$\mathbf E_{2A}$$ of the negative charge $$q_2$$ points towards $$q_2$$ (also towards the right). Each has magnitude

$$E_{1A} = E_{2A} = \frac{k|q|}{r^2} = \frac{9 \times 10^9 \times 10^{-8}}{(0.05)^2} = 3.6 \times 10^4 \, \mathrm{N\,C^{-1}}$$

As they point the same way, they add:

$$E_A = E_{1A} + E_{2A} = 7.2 \times 10^4 \, \mathrm{N\,C^{-1}}, \;\; \text{directed towards the right.}$$

At point B (on the line, $$0.05 \, \mathrm{m}$$ to the left of $$q_1$$, hence $$0.15 \, \mathrm{m}$$ from $$q_2$$):

$$E_{1B} = \frac{9 \times 10^9 \times 10^{-8}}{(0.05)^2} = 3.6 \times 10^4 \, \mathrm{N\,C^{-1}} \;\; (\text{towards the left, away from } q_1)$$

$$E_{2B} = \frac{9 \times 10^9 \times 10^{-8}}{(0.15)^2} = 4 \times 10^3 \, \mathrm{N\,C^{-1}} \;\; (\text{towards the right, towards } q_2)$$

These oppose each other, so

$$E_B = E_{1B} - E_{2B} = 3.6 \times 10^4 - 0.4 \times 10^4 = 3.2 \times 10^4 \, \mathrm{N\,C^{-1}}, \;\; \text{directed towards the left.}$$

At point C (the apex of an equilateral triangle, $$0.1 \, \mathrm{m}$$ from each charge):

$$E_{1C} = E_{2C} = \frac{9 \times 10^9 \times 10^{-8}}{(0.1)^2} = 9 \times 10^3 \, \mathrm{N\,C^{-1}}$$

$$\mathbf E_{1C}$$ points away from $$q_1$$ and $$\mathbf E_{2C}$$ points towards $$q_2$$; each makes an angle of $$60^\circ$$ with the line $$q_1q_2$$. The components perpendicular to that line cancel, while the components parallel to it (each equal to $$E\cos 60^\circ$$) add:

$$E_C = E_{1C}\cos 60^\circ + E_{2C}\cos 60^\circ = 2 \times 9 \times 10^3 \times \tfrac12 = 9 \times 10^3 \, \mathrm{N\,C^{-1}}$$

directed parallel to the line, towards the right.

The dipole has charges $$\pm 10 \, \mu\mathrm{C} = \pm 10^{-5} \, \mathrm{C}$$ separated by $$2a = 5.0 \, \mathrm{mm}$$, so $$a = 2.5 \times 10^{-3} \, \mathrm{m}$$. The dipole moment magnitude is $$p = q(2a) = 10^{-5} \times 5 \times 10^{-3} = 5 \times 10^{-8} \, \mathrm{C\,m}$$. Use $$k = 9 \times 10^9$$.

(a) Point P on the axis, $$r = 15 \, \mathrm{cm} = 0.15 \, \mathrm{m}$$ from O, on the side of $$+q$$.

P is at distance $$r - a = 0.1475 \, \mathrm{m}$$ from $$+q$$ and $$r + a = 0.1525 \, \mathrm{m}$$ from $$-q$$.

$$E_{+q} = \frac{kq}{(r-a)^2} = \frac{9 \times 10^9 \times 10^{-5}}{(0.1475)^2} \approx 4.13 \times 10^6 \, \mathrm{N\,C^{-1}} \;\; (\text{along OP})$$

$$E_{-q} = \frac{kq}{(r+a)^2} = \frac{9 \times 10^9 \times 10^{-5}}{(0.1525)^2} \approx 3.86 \times 10^6 \, \mathrm{N\,C^{-1}} \;\; (\text{towards O, opposite to OP})$$

The resultant points along OP (the direction of $$\mathbf p$$):

$$E_P = E_{+q} - E_{-q} \approx 4.13 \times 10^6 - 3.86 \times 10^6 = 2.7 \times 10^5 \, \mathrm{N\,C^{-1}}$$

Since $$r \gg a$$, this agrees with the axial dipole formula $$E = \dfrac{2kp}{r^3} = \dfrac{2 \times 9 \times 10^9 \times 5 \times 10^{-8}}{(0.15)^3} \approx 2.6 \times 10^5 \, \mathrm{N\,C^{-1}}$$.

(b) Point Q on the equatorial line, $$r = 0.15 \, \mathrm{m}$$ from O.

Q is at distance $$\sqrt{r^2 + a^2}$$ from each charge. The field magnitude due to each charge is

$$E_{+q} = E_{-q} = \frac{kq}{r^2 + a^2} = \frac{9 \times 10^9 \times 10^{-5}}{(0.15)^2 + (0.0025)^2} \approx 3.99 \times 10^6 \, \mathrm{N\,C^{-1}}$$

The components perpendicular to the dipole axis cancel; the components along the axis (each multiplied by $$\cos\theta = a/\sqrt{r^2+a^2}$$) add:

$$E_Q = 2E_{+q}\cos\theta = 2 \times 3.99 \times 10^6 \times \frac{0.0025}{0.15} \approx 1.33 \times 10^5 \, \mathrm{N\,C^{-1}}$$

directed antiparallel to $$\mathbf p$$. This matches the equatorial dipole formula $$E = \dfrac{kp}{r^3} = \dfrac{9 \times 10^9 \times 5 \times 10^{-8}}{(0.15)^3} \approx 1.33 \times 10^5 \, \mathrm{N\,C^{-1}}$$.

The field has only an $$x$$-component, $$E_x = \alpha x^{1/2}$$. For the four faces of the cube perpendicular to the $$y$$ and $$z$$ axes, $$\mathbf E$$ is parallel to the surface ($$\theta = 90^\circ$$), so the flux through them is zero. Only the two faces perpendicular to the $$x$$-axis contribute.

Place the left face at $$x = a$$ and the right face at $$x = 2a$$. The field magnitudes there are

$$E_L = \alpha a^{1/2}, \qquad E_R = \alpha (2a)^{1/2}$$

(a) Flux through the cube. The outward normal of the left face points along $$-x$$, so $$\theta = 180^\circ$$:

$$\phi_L = E_L a^2 \cos 180^\circ = -E_L a^2 = -\alpha a^{1/2}\,a^2$$

The outward normal of the right face points along $$+x$$, so $$\theta = 0^\circ$$:

$$\phi_R = E_R a^2 \cos 0^\circ = +E_R a^2 = \alpha (2a)^{1/2}\,a^2$$

The net flux is

$$\phi = \phi_R + \phi_L = \alpha a^2\left[(2a)^{1/2} - a^{1/2}\right] = \alpha a^{5/2}\left(\sqrt2 - 1\right)$$

$$\phi = 800 \times (0.1)^{5/2} \times (\sqrt2 - 1) = 800 \times 3.16 \times 10^{-3} \times 0.414 \approx 1.05 \, \mathrm{N\,m^2\,C^{-1}}$$

(b) Charge within the cube. By Gauss's law $$\phi = q/\varepsilon_0$$, so

$$q = \varepsilon_0\,\phi = 8.854 \times 10^{-12} \times 1.05 \approx 9.27 \times 10^{-12} \, \mathrm{C}$$

The cylinder has radius $$5 \, \mathrm{cm} = 0.05 \, \mathrm{m}$$, so each flat face has area

$$S = \pi r^2 = \pi (0.05)^2 \approx 7.85 \times 10^{-3} \, \mathrm{m^2}$$

(a) Flux through each flat face.

The right face is at $$x = +10 \, \mathrm{cm}$$, where $$\mathbf E = 200\,\hat{\mathbf i} \, \mathrm{N/C}$$. Its outward normal also points along $$+\hat{\mathbf i}$$, so $$\mathbf E$$ and $$\Delta\mathbf S$$ are parallel:

$$\phi_R = \mathbf E \cdot \Delta\mathbf S = 200 \times S = 200 \times 7.85 \times 10^{-3} \approx +1.57 \, \mathrm{N\,m^2\,C^{-1}}$$

The left face is at $$x = -10 \, \mathrm{cm}$$, where $$\mathbf E = -200\,\hat{\mathbf i} \, \mathrm{N/C}$$. Its outward normal points along $$-\hat{\mathbf i}$$, so $$\mathbf E$$ and $$\Delta\mathbf S$$ are again parallel:

$$\phi_L = \mathbf E \cdot \Delta\mathbf S = (-200)\times(-S) \approx +1.57 \, \mathrm{N\,m^2\,C^{-1}}$$

So the net outward flux through each flat face is $$+1.57 \, \mathrm{N\,m^2\,C^{-1}}$$.

(b) Flux through the curved side. Everywhere on the side, $$\mathbf E$$ (along $$\pm\hat{\mathbf i}$$) is perpendicular to the outward normal (which points radially outward). Hence $$\mathbf E \cdot \Delta\mathbf S = 0$$ and the flux through the side is zero.

(c) Net outward flux through the cylinder.

$$\phi = \phi_R + \phi_L + \phi_{\text{side}} = 1.57 + 1.57 + 0 = 3.14 \, \mathrm{N\,m^2\,C^{-1}}$$

(d) Net charge inside. By Gauss's law,

$$q = \varepsilon_0\,\phi = 8.854 \times 10^{-12} \times 3.14 \approx 2.78 \times 10^{-11} \, \mathrm{C}$$

The atom has a point nucleus of charge $$+Ze$$ at the centre, surrounded by a uniform negative charge spread through a sphere of radius $$R$$. Since the atom is neutral, the total negative charge must be $$-Ze$$.

Negative charge density. The negative charge $$-Ze$$ fills a sphere of volume $$\tfrac43\pi R^3$$, so

$$\rho = \frac{-Ze}{\tfrac43\pi R^3} = -\frac{3Ze}{4\pi R^3}$$

By spherical symmetry, $$\mathbf E$$ is radial and depends only on $$r$$. Choose a spherical Gaussian surface of radius $$r$$ centred on the nucleus.

(i) Inside the distribution, $$r < R$$. The enclosed charge is the nuclear charge plus the negative charge within radius $$r$$:

$$q_{\text{enc}} = Ze + \rho\left(\frac{4}{3}\pi r^3\right) = Ze - \frac{3Ze}{4\pi R^3}\cdot\frac{4\pi r^3}{3} = Ze\left(1 - \frac{r^3}{R^3}\right)$$

Gauss's law $$E\,(4\pi r^2) = q_{\text{enc}}/\varepsilon_0$$ gives

$$E(r) = \frac{Ze}{4\pi\varepsilon_0}\left(\frac{1}{r^2} - \frac{r}{R^3}\right), \qquad r < R$$

directed radially outward.

(ii) Outside the atom, $$r > R$$. The Gaussian surface now encloses the whole atom, whose total charge is zero. Hence

$$E\,(4\pi r^2) = \frac{0}{\varepsilon_0} \;\;\Rightarrow\;\; E(r) = 0, \qquad r > R$$

At $$r = R$$, both expressions give $$E = 0$$, so the field is continuous there.

Both spheres are small, so they may be treated as point charges. By Coulomb's law,

$$F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2} = k\,\frac{q_1 q_2}{r^2}$$

with $$k = 9 \times 10^9 \, \mathrm{N\,m^2\,C^{-2}}$$, $$q_1 = 2 \times 10^{-7} \, \mathrm{C}$$, $$q_2 = 3 \times 10^{-7} \, \mathrm{C}$$ and $$r = 30 \, \mathrm{cm} = 0.30 \, \mathrm{m}$$.

$$F = \frac{9 \times 10^9 \times (2 \times 10^{-7}) \times (3 \times 10^{-7})}{(0.30)^2}$$

$$F = \frac{9 \times 10^9 \times 6 \times 10^{-14}}{0.09} = \frac{5.4 \times 10^{-4}}{0.09} = 6 \times 10^{-3} \, \mathrm{N}$$

Both charges are positive, so the force is repulsive.

Dimensional check. From Coulomb's law $$F = \dfrac{k e^2}{r^2}$$, we have $$k e^2 = F r^2$$, with dimensions

$$[k e^2] = [\text{force}][\text{length}]^2 = [\mathrm{M\,L\,T^{-2}}][\mathrm{L^2}] = [\mathrm{M\,L^3\,T^{-2}}]$$

From Newton's law of gravitation $$F = \dfrac{G m_e m_p}{r^2}$$, we have $$G m_e m_p = F r^2$$, with the same dimensions

$$[G m_e m_p] = [\text{force}][\text{length}]^2 = [\mathrm{M\,L^3\,T^{-2}}]$$

Therefore the ratio

$$\frac{k e^2}{G\, m_e m_p} = \frac{[\mathrm{M\,L^3\,T^{-2}}]}{[\mathrm{M\,L^3\,T^{-2}}]}$$

is dimensionless.

Numerical value. Using $$k = 9 \times 10^9 \, \mathrm{N\,m^2\,C^{-2}}$$, $$e = 1.6 \times 10^{-19} \, \mathrm{C}$$, $$G = 6.67 \times 10^{-11} \, \mathrm{N\,m^2\,kg^{-2}}$$, $$m_e = 9.11 \times 10^{-31} \, \mathrm{kg}$$ and $$m_p = 1.67 \times 10^{-27} \, \mathrm{kg}$$:

$$k e^2 = 9 \times 10^9 \times (1.6 \times 10^{-19})^2 = 2.3 \times 10^{-28}$$

$$G\, m_e m_p = 6.67 \times 10^{-11} \times 9.11 \times 10^{-31} \times 1.67 \times 10^{-27} \approx 1.0 \times 10^{-67}$$

$$\frac{k e^2}{G\, m_e m_p} = \frac{2.3 \times 10^{-28}}{1.0 \times 10^{-67}} \approx 2.3 \times 10^{39}$$

Significance. This ratio is the strength of the electrostatic force between an electron and a proton compared with the gravitational force between the same pair. Its huge value ($$\sim 10^{39}$$) shows that the electric force is overwhelmingly stronger than the gravitational force; gravity is utterly negligible at the atomic scale.

When a glass rod is rubbed with silk, no new charge is created. Some electrons are merely transferred from the glass rod to the silk cloth. The glass rod, having lost electrons, acquires a positive charge $$+q$$; the silk, having gained exactly those electrons, acquires an equal negative charge $$-q$$.

Before rubbing, both bodies were neutral, so the total charge of the isolated glass-silk system was

$$q_{\text{before}} = 0 + 0 = 0$$

After rubbing, the total charge is

$$q_{\text{after}} = (+q) + (-q) = 0$$

The total charge is unchanged. The charges appearing on the two bodies are always equal in magnitude and opposite in sign, because one body gains precisely what the other loses. This is exactly the statement of the law of conservation of charge: charge can be transferred from one body to another, but it can neither be created nor destroyed. The same reasoning applies to every pair of bodies that are rubbed together.

Label the square ABCD with side $$10 \, \mathrm{cm}$$. The corner charges are $$q_A = 2 \, \mu\mathrm{C}$$, $$q_B = -5 \, \mu\mathrm{C}$$, $$q_C = 2 \, \mu\mathrm{C}$$, $$q_D = -5 \, \mu\mathrm{C}$$, and a charge $$q_O = 1 \, \mu\mathrm{C}$$ sits at the centre O.

The centre O is equidistant from all four corners; this distance is half the diagonal:

$$r = \frac{1}{2}\times(\text{diagonal}) = \frac{1}{2}\times\left(\sqrt2 \times 0.10\right) = \frac{0.10}{\sqrt2} \, \mathrm{m}$$

Charges A and C: A and C lie at opposite ends of one diagonal. They carry equal charges ($$2 \, \mu\mathrm{C}$$ each) at equal distances $$r$$ from O. The force on $$q_O$$ due to A and the force due to C therefore have equal magnitude

$$F_A = F_C = k\,\frac{q_O\,q_A}{r^2}$$

but act along the same diagonal in opposite directions (one towards A, the other towards C). Hence they cancel exactly.

Charges B and D: Likewise, B and D lie at opposite ends of the other diagonal and carry equal charges ($$-5 \, \mu\mathrm{C}$$ each) at equal distances $$r$$. The forces they exert on $$q_O$$ are equal in magnitude and opposite in direction, so they too cancel.

Adding the two cancelling pairs:

$$\mathbf F_{\text{net}} = (\mathbf F_A + \mathbf F_C) + (\mathbf F_B + \mathbf F_D) = 0 + 0 = 0$$

The net force on the $$1 \, \mu\mathrm{C}$$ charge at the centre is zero.

Total charge. Charge is a scalar and adds algebraically:

$$q_{\text{total}} = q_A + q_B = (2.5 \times 10^{-7}) + (-2.5 \times 10^{-7}) = 0$$

Electric dipole moment. The two charges are equal and opposite, so they form an electric dipole. Their separation is

$$2a = 15 \, \mathrm{cm} + 15 \, \mathrm{cm} = 30 \, \mathrm{cm} = 0.30 \, \mathrm{m}$$

The magnitude of the dipole moment is

$$p = q \times 2a = (2.5 \times 10^{-7}) \times 0.30 = 7.5 \times 10^{-8} \, \mathrm{C\,m}$$

By convention the dipole moment vector points from the negative charge to the positive charge. Here the negative charge $$q_B$$ is at $$z = +15 \, \mathrm{cm}$$ and the positive charge $$q_A$$ is at $$z = -15 \, \mathrm{cm}$$. So $$\mathbf p$$ points from B to A, i.e. along the negative $$z$$-direction.

The torque on a dipole of moment $$p$$ placed in a uniform field $$E$$ is $$\boldsymbol\tau = \mathbf p \times \mathbf E$$, whose magnitude is

$$\tau = pE\sin\theta$$

where $$\theta$$ is the angle between $$\mathbf p$$ and $$\mathbf E$$. Here $$p = 4 \times 10^{-9} \, \mathrm{C\,m}$$, $$E = 5 \times 10^4 \, \mathrm{N\,C^{-1}}$$ and $$\theta = 30^\circ$$:

$$\tau = (4 \times 10^{-9}) \times (5 \times 10^4) \times \sin 30^\circ$$

$$\tau = (4 \times 10^{-9}) \times (5 \times 10^4) \times 0.5$$

$$\tau = 10 \times 10^{-5} = 1 \times 10^{-4} \, \mathrm{N\,m}$$

In Fig. 1.30 the upper plate is positive and the lower plate is negative, so the uniform electric field between the plates points downward (from the positive plate towards the negative plate).

Signs of the charges. The force on a charge is $$\mathbf F = q\mathbf E$$.

• Particles 1 and 2 are deflected upward, towards the positive plate. Their force is opposite to $$\mathbf E$$, so they must carry negative charge.
• Particle 3 is deflected downward, towards the negative plate. Its force is along $$\mathbf E$$, so it carries positive charge.

Highest charge-to-mass ratio. While crossing the field region (the same horizontal length, at roughly the same speed), a particle is displaced sideways by

$$y = \frac{1}{2}at^2 = \frac{1}{2}\left(\frac{qE}{m}\right)t^2$$

so the deflection is proportional to the charge-to-mass ratio $$q/m$$. Particle 3 shows the largest deflection, so it has the highest charge-to-mass ratio.

The field $$\mathbf E = 3 \times 10^3\,\hat{\mathbf i} \, \mathrm{N/C}$$ is uniform. The cube is oriented with its faces parallel to the coordinate planes.

Consider the flux face by face:

• The four faces parallel to the $$x$$-axis have their outward normals perpendicular to $$\mathbf E$$, so the flux through each of them is zero.
• For the two faces perpendicular to the $$x$$-axis, the field lines enter through one face and leave through the opposite face. The flux entering one face is $$-EA$$ and the flux leaving the opposite face is $$+EA$$.

Adding all contributions:

$$\phi_{\text{net}} = (+EA) + (-EA) + 0 + 0 + 0 + 0 = 0$$

The net flux is zero. This also follows directly from Gauss's law: a uniform field means no charge is enclosed inside the cube, so $$\phi = q_{\text{enc}}/\varepsilon_0 = 0$$.

Following the hint, imagine the square as one face of a cube of edge $$10 \, \mathrm{cm}$$. Since the charge $$+10 \, \mu\mathrm{C}$$ is $$5 \, \mathrm{cm}$$ directly above the centre of the square, it sits exactly at the centre of this cube.

By Gauss's law, the total flux through the whole closed cubical surface is

$$\phi_{\text{total}} = \frac{q}{\varepsilon_0}$$

By symmetry, a charge at the centre sends an equal flux through each of the $$6$$ identical faces. So the flux through one face (our square) is

$$\phi_{\text{square}} = \frac{1}{6}\cdot\frac{q}{\varepsilon_0}$$

Substituting $$q = 10 \times 10^{-6} \, \mathrm{C}$$ and $$\varepsilon_0 = 8.854 \times 10^{-12} \, \mathrm{C^2\,N^{-1}\,m^{-2}}$$:

$$\phi_{\text{square}} = \frac{10 \times 10^{-6}}{6 \times 8.854 \times 10^{-12}} = \frac{10 \times 10^{-6}}{5.31 \times 10^{-11}}$$

$$\phi_{\text{square}} \approx 1.88 \times 10^5 \, \mathrm{N\,m^2\,C^{-1}}$$

By Gauss's law, the net electric flux through any closed surface depends only on the total charge enclosed, not on the size or shape of the surface:

$$\phi = \frac{q}{\varepsilon_0}$$

The point charge $$q = 2.0 \, \mu\mathrm{C} = 2.0 \times 10^{-6} \, \mathrm{C}$$ is at the centre of the cube, so all of it is enclosed:

$$\phi = \frac{2.0 \times 10^{-6}}{8.854 \times 10^{-12}}$$

$$\phi \approx 2.26 \times 10^5 \, \mathrm{N\,m^2\,C^{-1}}$$

Note that the answer does not depend on the $$9.0 \, \mathrm{cm}$$ edge length - the flux would be the same for a cube of any size.

For any point outside a charged conducting sphere, the sphere behaves as if its entire charge $$q$$ were concentrated at its centre. So at a distance $$r$$ from the centre,

$$E = \frac{1}{4\pi\varepsilon_0}\frac{q}{r^2} = k\,\frac{q}{r^2}$$

Here $$r = 20 \, \mathrm{cm} = 0.20 \, \mathrm{m}$$ (which lies outside the sphere of radius $$10 \, \mathrm{cm}$$) and $$E = 1.5 \times 10^3 \, \mathrm{N/C}$$. Solving for the charge magnitude:

$$q = \frac{E\,r^2}{k} = \frac{(1.5 \times 10^3) \times (0.20)^2}{9 \times 10^9}$$

$$q = \frac{(1.5 \times 10^3) \times 0.04}{9 \times 10^9} = \frac{60}{9 \times 10^9}$$

$$q \approx 6.67 \times 10^{-9} \, \mathrm{C}$$

The field points radially inward, which means the charge is negative. Therefore

$$q \approx -6.67 \times 10^{-9} \, \mathrm{C} = -6.67 \, \mathrm{nC}$$

The electric field at a perpendicular distance $$r$$ from an infinitely long line charge of linear charge density $$\lambda$$ is

$$E = \frac{\lambda}{2\pi\varepsilon_0\,r}$$

Solving for $$\lambda$$:

$$\lambda = 2\pi\varepsilon_0\,r\,E$$

It is convenient to write $$\dfrac{1}{2\pi\varepsilon_0} = 2k$$ with $$k = 9 \times 10^9$$, so that $$E = \dfrac{2k\lambda}{r}$$ and

$$\lambda = \frac{E\,r}{2k} = \frac{(9 \times 10^4) \times (0.02)}{2 \times (9 \times 10^9)}$$

$$\lambda = \frac{1.8 \times 10^3}{1.8 \times 10^{10}} = 1 \times 10^{-7} \, \mathrm{C/m}$$

So the linear charge density is $$10^{-7} \, \mathrm{C/m} = 0.1 \, \mu\mathrm{C/m}$$.