Electrostatics

Like charges repel and unlike charges attract. Two positive charges are alike, so the force between them is a repulsion, pushing them apart.

Charging by friction transfers electrons, never protons. A negatively charged rod has gained extra electrons; the cloth, having lost them, is left equally positive. Charge is conserved – only moved.

Metals contain free (delocalised) electrons that can drift through the material, so charge moves easily. In an insulator the electrons are tightly bound, so charge tends to stay where it is placed.

An electric field line shows the direction of the force a positive test charge would feel. Lines therefore start on positive charges and end on negative ones.

Coulomb's law is $F = \dfrac{kq_1q_2}{r^2}$: the force grows with the product of the two charges and falls off as the square of the distance between them.

A positive test charge is pushed away from a positive source charge, so the field lines radiate straight outward in every direction. Around a negative charge they point inward instead.

$F = \dfrac{(9.0\times10^{9})(3\times10^{-6})(2\times10^{-6})}{(0.30)^2} = \dfrac{9.0\times10^{9}\times 6\times10^{-12}}{0.09} = 0.60\text{ N}$.

The force follows an inverse-square law, $F \propto \dfrac{1}{r^2}$. Doubling $r$ multiplies the force by $\left(\tfrac{1}{2}\right)^2 = \tfrac{1}{4}$. (Thinking it just halves is the common trap.)

The force is proportional to the product $q_1q_2$. Tripling just one charge multiplies the product – and so the force – by 3. (Tripling both charges would give a factor of $9$.)

$E = \dfrac{F}{q} = \dfrac{0.020}{4\times10^{-6}} = 5000\text{ N/C}$.

$E = \dfrac{V}{d} = \dfrac{200}{0.040} = 5000\text{ V/m}$. Between parallel plates the field is uniform, so it has this same strength everywhere between them.

Between parallel plates the field lines are evenly spaced and parallel, so the field has the same strength (and direction) at every point between them, apart from slight fringing at the edges. That uniformity is exactly why a single value $E = \dfrac{V}{d}$ describes the whole region.

$E = \dfrac{kq}{r^2} = \dfrac{(9.0\times10^{9})(5\times10^{-6})}{(0.30)^2} = \dfrac{4.5\times10^{4}}{0.09} = 5\times10^{5}\text{ N/C}$.

The spacing of field lines represents the field strength: closely spaced lines (as at A) mean a strong field, widely spaced lines (as at B) a weak one. That is why the lines crowd together near a charge and spread out far away.

The nearby negative rod repels electrons in the sphere; with the sphere grounded, those electrons flow away to earth. Removing the ground first, then the rod, leaves the sphere short of electrons – so it is left positively charged, the opposite sign to the rod.

Between the two charges their fields point in opposite directions. Because the charges are equal, the fields are equal in size only at the point equally far from both – the midpoint – where they cancel exactly.

The electric force is conservative, so the work is $W = qV$ – fixed entirely by the potential difference between the start and end points. Every path between the same two points requires the same work, just as lifting a mass to a given height takes the same energy by any route.

The distance triples ($0.60 / 0.20 = 3$), and $F \propto \dfrac{1}{r^2}$, so the force drops by $3^2 = 9$: $\dfrac{0.90}{9} = 0.10\text{ N}$. (Dividing by just $3$ gives the $0.30\text{ N}$ trap.)

Each charge is $0.10\text{ m}$ from the midpoint, giving $E = \dfrac{kq}{r^2} = \dfrac{(9.0\times10^{9})(2\times10^{-6})}{(0.10)^2} = 1.8\times10^{6}\text{ N/C}$ each. At the midpoint both fields point the same way (away from the $+$, toward the $-$), so they add: $3.6\times10^{6}\text{ N/C}$. (For two like charges the fields would instead cancel – the $0$ trap.)

The field is $E = \dfrac{V}{d} = \dfrac{100}{0.020} = 5000\text{ N/C}$. Then $F = qE = (5\times10^{-6})(5000) = 0.025\text{ N}$.

The $+4\,\mu\text{C}$ on the left repels the middle charge to the right; the $-4\,\mu\text{C}$ on the right attracts it to the right. Both forces point the same way and add. Each is $F = \dfrac{(9.0\times10^{9})(2\times10^{-6})(4\times10^{-6})}{(0.20)^2} = 1.8\text{ N}$, so the net force is $3.6\text{ N}$. (Assuming the $+4$ and $-4$ simply cancel gives the $0$ trap.)

For the fields to cancel, the point must be nearer the weaker charge, so that its smaller field – measured over a shorter distance – can match the stronger charge's field measured over a longer one. So the zero-field point sits closer to the $+1\,\mu\text{C}$ charge.

For any everyday charge the Coulomb force is enormously stronger than gravity – for two protons the electric repulsion is around $10^{36}$ times their gravitational attraction. Gravity only wins out for large, essentially neutral masses like planets and stars.

Between two unlike charges both fields point the same way (toward the negative charge), so they can never cancel there. The zero must lie outside the pair, beyond the weaker $-1\,\mu\text{C}$ charge – close enough to it that its field can grow to match the more distant $+4\,\mu\text{C}$ charge's field.

The ball is in equilibrium under three forces: its weight (down), the thread tension (along the thread) and the horizontal electric force. Resolving the tension, its horizontal and vertical parts must balance the electric force and the weight, so $\tan 30^\circ = \dfrac{F}{W}$. Then $F = W\tan 30^\circ = (0.020)(0.577) = 0.012\text{ N}$. (Using $W\sin 30^\circ$ or $W\cos 30^\circ$ gives the other traps.)