Charging by induction involves bring a charged rod near to the metal cap. The charged rad attracts charge of opposite polarity to the cap and repels charge of the same polarity to the leaf and stem. The cap is then earthed and the charged rod withdrawn leaving the charge on leaf and stem to redistribute itself over the cap, stem and leaf.

When a gold leaf oscilloscope is charged by contact, as below, a charged rod actually makes contact with the cap, allowing the charge to pass onto the cap and down to the stem and leaf. The oscilloscope is then withdrawn, leaving behind a net charge of the same polarity as the charged rod.

]]>Atoms are neutral. They contain equal numbers of positively charged protons and negatively charged electrons. The charge on electrons and protons is equal in magnitude, so overall the atom is neutral.

The Universe as a whole is probably neutral, and so is every large body in it, or very nearly so. The electric force between an electron and proton is abouttimes larger than the force of gravity, so if any large body is not electrically neutral, the charge on it must be very small or it would disturb the obit of that body.

Though charge can move from place to place, the amount of charge gained by one body equals the amount of charge lost by another. The net amount of charge remains the same. For example when you brush your hair, electrons move between your hair and the brush. The charge lost by the brush will be the charge gained by your hair. It can easily be shown that both are charged. Your hair will be attracted to the brush because for example, your hair is negatively charged and the brush is positively charged. Your hair stand up because the strands, being of like charge, repel each other. If you bring the brush close to a running tap, the water will be attracted to the brush.

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Wherever electric charges move an electric current is created. Usually only electrons carry electric charge. This is the case for all solids. Positively charged atoms are foxed in positions and only electrons are able to move. In a gas or solid, atoms may sometimes become ionised or charged - usually positive - and are then able to carry charge and current.

The single characteristic that separates metals from other materials is the presence of free electrons. Free electrons are able to move throughout the materials and so they are free to carry electric charge - being negatively charged - and electric current. Electrons, being negative, are repelled by negative charges and attracted by positive charges. Because of a historical accident, we consider the flow of electric current as being from positive to negative and call it the 'conventional current' - and the electric field as being in the direction of the force on a positive charge.]]>

Like forces repel, opposite forces attract.

This means that two positive forces repel each other, two negative forces repel each other and positive forces are attracted to negative forces and vice versa.

The force always acts along the line of force joining the two charges, and the forces constitute a Newton's third law force pair, so they are equal in magnitude and opposite in direction.

All of these are expressed by Coulomb's Law, illustrated below.

If two or more charges are brought close together, the net force on each charge is the sum of the force due to all the other charges. Force is a vector, and each force will act in a particular direction. We can add these forces to obtain the net charge.

Coulomb's Law is an inverse square law, so the electric force decreases rapidly with increasing distance.

]]>Current flows through an object as a result of a potential difference across the object. The source of the potential difference may be a power supply, battery, thermocouple or any device capable of creating a voltage.

Though current is defined as the flow of electric charge, the convention has been adopted that current flows always from positive to negative terminals. This would be the path followed by charge carriers if they were positive. This is not usually the path of the actual charge carriers, which are usually negatively charged electrons. These flow from the negative to the positive terminals. The convention was adopted before the nature of electric charge or current were known.

Although current can flow in liquids, gases, or in fact in outer space, in most of the situations in which we are concerned, currents will be flowing in solids, specifically in wires in electric circuits.

In a wire, negatively charged electrons can move freely through a lattice consisting of positively charged metal ions. As they move, the electrons interact with the ions in the lattice, repeatedly colliding with them and giving up some of their energy to them. The lattice ions are made to vibrate about their mean position. This vibrational energy is heat energy, so that when a current flows the metal heats up. The electrons rapidly attain a steady speed throughout the wire, called the drift velocity.

]]>A electric charge or combination of electric charges produces an electric field – that is, a force is exerted on any electric charge that comes close to these charges. If a 'test charge'– an infinitesimally small charge, so that the distribution of charges is not disturbed and the electric field does not change -is introduced, the force experienced by the test charge will be the sum of the forces exerted by each individual charge in the distribution. We can use Coulomb's Law to write

whereis the unit vector from chargeto the test charge

Then

Dividing bygives

The electric field at the position ofdue to the chargeisso we can also write

This means that we can find the field due to each charge as a vector and add them.

The technical term is linear – the electric field and the electric force are both linear in the charges

]]>Some of the voltage battery is missing – the 'missing volts'. The reason is that the battery itself has some resistance, and some of the battery voltage – the emf or electromotive force, actually the energy per unit charge supplied by the battery – must be used to drive the current through the battery.

Alternatively we may say that the power made available to the external circuit – the resistor – is less than the power supplied by the chemical reaction inside the battery, with the difference being that power needed to drive the current against the internal resistance of the battery.

There is an equation to model the situation. If the emf of the battery is E and a current I flows through the resistance R, then the voltage across this resistance is IR and the lost volts isFor the circuit above right, the current through the batter isso if the internal resistance of the battery isthe lost volts isand we can write(1)

We can find the emfof a battery and it's internal resistanceusing the circuit below.

By varying the resistance, we can vary the voltage and current. Write (1) aswhere the voltage across the variable resistor. Thenwhich is of the form Values ofagainstcan be plotted. The gradient of the graph isand the intercept is

]]>If the field is uniform the force will be constant, since

The change in electrical potential energy is given bywhereis the distance moved in the direction of the field as shown below.

Alternatively we can writewhere as beforeis the distance moved and nowis the change in potential as the carge q moves from the start point to the end point. Then the change in electrical potential energy is

This change in potential energy will result in a change in speed of the particle. A fall in electrical potential energy will result in an increase in kinetic energy because total energy is conserved. If the particle starts from rest then, it will move in the direction of the force on it.

Fall in electrical potential energy = increase in kinetic energy

]]>Notice that the flux or field lines are at right angle to the equipotential surfaces wherever they meet. This is in fact true in general for equipotentials and flux lines, a consequence of the fact that the force on a particle acts in the direction of the greatest rate of change of potential, which is always perpendicular to the equipotential. The flux lines of course, indicate the direction of the electric field, and the direction of the force on a positive charge.

The closer together the equipotentials, the stronger the electric field, since the gradient (and force) is most rapid where the equipotentials are closest. For this reason, the relative strength of an electric field at different points can be estimated by considering the equipotentials.

Equipotentials are always closed surfaces around a distribution of charge, though they may take different forms at different distances from the charges, while always smoothly deforming. The surface of a conductor is always an equipotential, since charge is free to move on the surface so as to equalise the potential everywhere. The surface being an equipotential, the electrical field is always perpendicular to the surface.

]]>The force on a charge {jatex options:inline}Q{/jatex} at point A due to a point charge {jatex options:inline}q_B{/jatex} at a point B is {jatex options:inline}\mathbf{F}_{BA}=\frac{1}{4 \pi \epsilon_0} \frac{q_B Q}{r^2_{AB}} \mathbf{e}_{BA}{/jatex} where {jatex options:inline}r_{AB}{/jatex} is the distance from A to B, {jatex options:inline}\epsilon_0 = 8.854 \times 10^{-12} F/m{/jatex} and {jatex options:inline}\mathbf{e}_{BA}{/jatex} is the unit vector from B to A.

Suppose then that we have five charges as shown.

Each gives rise to a force, but the overall force is

{jatex options:inline}\begin{equation} \begin{aligned} \mathbf{F} &= \mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3+\mathbf{F}_4+\mathbf{F}_5 \\ &= \frac{1}{4 \pi \epsilon_0} \frac{q_1Q}{d^2_1}\mathbf{e}_{1A}+ \frac{1}{4 \pi \epsilon_0} \frac{q_2Q}{d^2_2}\mathbf{e}_{2A}+ \frac{1}{4 \pi \epsilon_0} \frac{q_3Q}{d^2_3}\mathbf{e}_{3A} + \frac{1}{4 \pi \epsilon_0} \frac{q_4Q}{d^2_4}\mathbf{e}_{4A} + \frac{1}{4 \pi \epsilon_0} \frac{q_5Q}{d^2_5}\mathbf{e}_{5A} \\ &= \frac{Q}{4 \pi \epsilon_0} (\frac{q_1}{d^2_1}\mathbf{e}_{1A} + \frac{q_2}{d^2_2}\mathbf{e}_{2A} + \frac{q_3}{d^2_3}\mathbf{e}_{3A} + \frac{q_4}{d^2_4} \mathbf{e}_{4A} + \frac{q_5}{d^2_5} \mathbf{e}_{5A}) \end{aligned} \end{equation} {/jatex}]]>