Chapter 1: Electric Charges and Fields (Electrostatics)

Given: $q_1=2\times 10^{-7}$ C, $q_2=3\times 10^{-7}$ C, $r=30$ cm $=0.30$ m.

$$F=\frac{k{q}_1{q}_2}{r^2}=\frac{9\times 10^9\times 2\times 10^{-7}\times 3\times 10^{-7}}{(0.30{)}^2}=\frac{5.4\times 10^{-4}}{0.09}=6\times 10^{-3}\text{N}$$

Both charges are positive, so the force is repulsive.

$F=6\times 10^{-3}$ N, repulsive.

(a) $F=\frac{k|q_1{q}_2|}{r^2}$, so

$$r^2=\frac{k|q_1{q}_2|}{F}=\frac{9\times 10^9\times 0.4\times 10^{-6}\times 0.8\times 10^{-6}}{0.2}=\frac{2.88\times 10^{-3}}{0.2}=0.0144{\text{m}}^2$$

$r=\sqrt{0.0144}=0.12$ m $=12$ cm.

(b) By Newton's third law, the second sphere feels a force of the same magnitude, 0.2 N, in the opposite direction. The charges are unlike, so the force is attractive: it pulls the second sphere towards the first.

(a) 0.12 m (12 cm). (b) 0.2 N, attractive (towards the first sphere).

Dimensions. $\frac{k{e}^2}{r^2}$ is a force (Coulomb's law) and $\frac{G{m}_e{m}_p}{r^2}$ is also a force (Newton's law of gravitation). So

$$\frac{k{e}^2}{G{m}_e{m}_p}=\frac{k{e}^2/r^2}{G{m}_e{m}_p/r^2}=\frac{[\text{force}]}{[\text{force}]}$$

which has no dimensions.

Value. With $k=9\times 10^9$, $e=1.6\times 10^{-19}$ C, $G=6.67\times 10^{-11}$ N m² kg⁻², $m_e=9.1\times 10^{-31}$ kg, $m_p=1.67\times 10^{-27}$ kg:

$$\frac{k{e}^2}{G{m}_e{m}_p}=\frac{9\times 10^9\times (1.6\times 10^{-19}{)}^2}{6.67\times 10^{-11}\times 9.1\times 10^{-31}\times 1.67\times 10^{-27}}=\frac{2.30\times 10^{-28}}{1.01\times 10^{-67}}\approx 2.3\times 10^{39}$$

Meaning. It is the ratio of the electric force to the gravitational force between an electron and a proton at the same distance. The electric force is about $10^{39}$ times stronger, so gravity can be ignored inside atoms.

Dimensionless; about $2.3\times 10^{39}$: the electric attraction between an electron and a proton is about $10^{39}$ times their gravitational attraction.

(a) Charge exists only in whole-number multiples of the basic charge $e$: $q=ne$, where $n=0,\pm 1,\pm 2,\dots$ and $e=1.6\times 10^{-19}$ C. A body gains or loses charge only by gaining or losing whole electrons, so a charge like $1.5e$ is not possible.

(b) Large-scale charges are huge compared with $e$. For example, 1 μC contains about $6\times 10^{12}$ electronic charges. A step of one $e$ in such a charge is far too small to notice, so the charge behaves as if it were continuous.

(a) Every charge is $q=ne$ with $n$ an integer. (b) Everyday charges contain so many units of $e$ that the steps of $e$ are too small to matter.

Before rubbing, the rod and the cloth are both neutral (total charge zero). Rubbing does not create charge; it only transfers electrons from the glass to the silk. The glass loses electrons and becomes positive ($+q$); the silk gains the same electrons and becomes negative ($-q$). The total charge is still $+q+(-q)=0$, exactly as before. So charge is only moved from one body to the other, never created or destroyed, which is the law of conservation of charge.

Rubbing only transfers electrons; the two bodies get equal and opposite charges, so the total stays zero.

The centre O is at the same distance from all four corners.

• A and C are opposite corners with equal charges (+2 μC). They repel the 1 μC charge with equal forces in exactly opposite directions, so these two forces cancel.
• B and D are the other pair of opposite corners with equal charges (−5 μC). They attract the 1 μC charge with equal forces in opposite directions, so these also cancel.

So the net force is zero.

Zero: the forces from each pair of opposite corners cancel.

(a) A field line shows the path a small free positive test charge would follow. The field exists at every point between the charges, and the test charge moves smoothly from point to point; it cannot jump across a gap. So the line has no breaks.

(b) The tangent to a field line at a point gives the direction of the electric field there. If two lines crossed, there would be two tangents, meaning two directions of the field at one point. The field at a point has only one direction, so field lines never cross.

(a) The field exists at every point, so the path of a test charge is continuous. (b) A crossing would give two field directions at one point, which is impossible.

(a) O is $r=10$ cm $=0.1$ m from each charge.

Field due to A: $E_A=\frac{k{q}_A}{r^2}=\frac{9\times 10^9\times 3\times 10^{-6}}{(0.1{)}^2}=2.7\times 10^6$ N/C, directed away from A, that is towards B.

Field due to B: same size, $E_B=2.7\times 10^6$ N/C. B is negative, so its field points towards B.

Both point the same way (from A towards B), so they add:

$$E=E_A+E_B=5.4\times 10^6\text{N/C, along AB (towards B)}$$

(b) $F=qE=1.5\times 10^{-9}\times 5.4\times 10^6=8.1\times 10^{-3}$ N. The test charge is negative, so the force is opposite to $E$: towards A.

(a) $5.4\times 10^6$ N/C along OB. (b) $8.1\times 10^{-3}$ N towards A.

Total charge $=2.5\times 10^{-7}+(-2.5\times 10^{-7})=0$.

The separation is $2a=15+15=30$ cm $=0.30$ m, so

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

The dipole moment points from the negative charge to the positive charge: from B (z = +15 cm) to A (z = −15 cm), that is along the negative z-axis.

Total charge 0; $p=7.5\times 10^{-8}$ C m along the negative z-direction.

$$\tau =pE\sin \theta =4\times 10^{-9}\times 5\times 10^4\times \sin 30^{\circ }=2\times 10^{-4}\times 0.5=1\times 10^{-4}\text{N m}$$

$\tau =10^{-4}$ N m.

(a) $n=\frac{q}{e}=\frac{3\times 10^{-7}}{1.6\times 10^{-19}}\approx 1.9\times 10^{12}$ electrons. The polythene became negative, so the electrons moved from the wool to the polythene.

(b) Yes, a very small amount, because electrons have mass:

$$m=n{m}_e=1.875\times 10^{12}\times 9.1\times 10^{-31}\approx 1.7\times 10^{-18}\text{kg}$$

This is far too small to measure.

(a) About $2\times 10^{12}$ electrons, from wool to polythene. (b) Yes, but only about $1.7\times 10^{-18}$ kg, which is negligible.

(a) $r=0.5$ m:

$$F=\frac{k{q}^2}{r^2}=\frac{9\times 10^9\times (6.5\times 10^{-7}{)}^2}{(0.5{)}^2}=\frac{3.80\times 10^{-3}}{0.25}\approx 1.5\times 10^{-2}\text{N}$$

(b) Doubling each charge multiplies $q_1{q}_2$ by 4; halving $r$ divides $r^2$ by 4, which multiplies $F$ by 4. So $F^{\prime }=16F=16\times 1.52\times 10^{-2}\approx 0.24$ N.

(a) $1.5\times 10^{-2}$ N. (b) 16 times larger, about 0.24 N.

A charged particle in the field between the plates bends towards the plate of opposite sign.

• Particles 1 and 2 bend towards the positive plate, so they are negatively charged.
• Particle 3 bends towards the negative plate, so it is positively charged.

The sideways deflection is $y=\frac{1}{2}\frac{qE}{m}{\left(\frac{x}{v}\right)}^2$, which is proportional to $q/m$ for the same speed. Particle 3 is deflected the most, so it has the highest charge-to-mass ratio.

1 and 2 are negative, 3 is positive; particle 3 has the highest $q/m$.

Area $A=(0.1{)}^2=0.01$ m².

(a) A plane parallel to the yz plane has its normal along the x-axis, the same direction as $\vec{E}$ ($\theta =0^{\circ }$):

$\varphi =EA\cos 0^{\circ }=3\times 10^3\times 0.01=30$ N m²/C.

(b) $\varphi =EA\cos 60^{\circ }=30\times 0.5=15$ N m²/C.

(a) 30 N m²/C. (b) 15 N m²/C.

The field lines that enter one face of the cube leave through the opposite face; the flux in is equal to the flux out. Also, the cube encloses no charge, so by Gauss's law $\varphi =q/{\epsilon }_0=0$.

Zero.

(a) By Gauss's law, $q={\epsilon }_0\varphi =8.854\times 10^{-12}\times 8.0\times 10^3\approx 7.1\times 10^{-8}$ C $=0.07$ μC.

(b) No. Zero flux only means the net enclosed charge is zero. The box could contain equal positive and negative charges (for example a dipole).

(a) About $7.1\times 10^{-8}$ C. (b) No; only the net charge is zero, equal + and − charges could be inside.

Imagine a cube of edge 10 cm with the charge at its centre (5 cm from each face). The square is one face of this cube. By symmetry the total flux $q/{\epsilon }_0$ divides equally among the 6 faces:

$$\varphi =\frac{q}{6{\epsilon }_0}=\frac{10\times 10^{-6}}{6\times 8.854\times 10^{-12}}\approx 1.9\times 10^5{\text{N m}}^2/\text{C}$$

$\varphi \approx 1.9\times 10^5$ N m²/C.

The size of the cube does not matter; only the enclosed charge does:

$$\varphi =\frac{q}{{\epsilon }_0}=\frac{2.0\times 10^{-6}}{8.854\times 10^{-12}}\approx 2.3\times 10^5{\text{N m}}^2/\text{C}$$

$\varphi \approx 2.3\times 10^5$ N m²/C.

(a) The flux depends only on the enclosed charge, not on the size of the surface. So it stays $-1.0\times 10^3$ N m²/C.

(b) $q={\epsilon }_0\varphi =8.854\times 10^{-12}\times (-1.0\times 10^3)\approx -8.9\times 10^{-9}$ C.

(a) Still $-1.0\times 10^3$ N m²/C. (b) $q\approx -8.9\times 10^{-9}$ C (−8.9 nC).

Outside a charged sphere the field is as if all the charge were at the centre: $E=\frac{k|q|}{r^2}$ with $r=0.20$ m.

$$|q|=\frac{E{r}^2}{k}=\frac{1.5\times 10^3\times (0.20{)}^2}{9\times 10^9}=\frac{60}{9\times 10^9}\approx 6.7\times 10^{-9}\text{C}$$

The field points inward, so the charge is negative: $q\approx -6.7\times 10^{-9}$ C.

$q\approx -6.7$ nC.

Radius $r=1.2$ m, $\sigma =80.0\times 10^{-6}$ C/m².

(a) $q=\sigma \times 4\pi r^2=80.0\times 10^{-6}\times 4\pi \times (1.2{)}^2=80.0\times 10^{-6}\times 18.1\approx 1.45\times 10^{-3}$ C.

(b) $\varphi =\frac{q}{{\epsilon }_0}=\frac{1.45\times 10^{-3}}{8.854\times 10^{-12}}\approx 1.6\times 10^8$ N m²/C.

(a) $1.45\times 10^{-3}$ C. (b) $1.6\times 10^8$ N m²/C.

For an infinite line charge, $E=\frac{\lambda }{2\pi {\epsilon }_{0}r}=\frac{2k\lambda }{r}$. So

$$\lambda =\frac{Er}{2k}=\frac{9\times 10^4\times 0.02}{2\times 9\times 10^9}=1\times 10^{-7}\text{C/m}$$

$\lambda =10^{-7}$ C/m (0.1 μC/m).

Each plate gives a field $\frac{\sigma }{2{\epsilon }_0}$. Outside the pair, the two fields are opposite and cancel; between the plates they point the same way and add.

(a) and (b): $E=0$ in both outer regions.

(c) Between the plates:

$$E=\frac{\sigma }{{\epsilon }_0}=\frac{17.0\times 10^{-22}}{8.854\times 10^{-12}}\approx 1.9\times 10^{-10}\text{N/C}$$

directed from the positive plate to the negative plate.

(a) 0. (b) 0. (c) $1.9\times 10^{-10}$ N/C, from the positive plate to the negative plate.