Charge and electric field

Positive and negative test charges are placed in turn at the same point. The sources are unchanged, yet the forces point in opposite directions. How can we describe the influence around the sources independently of the test object's properties?

When two objects are charged, one may gain an excess of electrons while the other loses electrons. Electrons carry negative charge: gaining electrons makes an object negative; losing them makes it positive. Electric charge q is measured in coulombs, C, and includes a sign. Ordinary charging redistributes charge; rubbing does not create it from nothing. Neutrality means positive and negative charges balance, not that charged particles are absent.

Draw an imaginary boundary around an electrically isolated system. If no charge crosses it, q_1+q_2+\ldots+q_n=\text{const}: the law of charge conservation. Indices label different objects and the sum is algebraic. Electron transfer can change each term substantially while preserving the total. For example, equal positive and negative charges leave a zero total. Grounding allows charge exchange, so check the chosen boundary.

Like charges repel and unlike charges attract. For stationary point charges, the force lies along the line joining them: this pair selects no other direction. The point model works when object sizes are negligible compared with their separation. Force magnitude depends on both charges, distance, and the medium. Draw the direction before calculating; the type of interaction then provides an independent check.

Charges placed in matter redistribute its own charges, so the medium changes the resulting interaction. Let r be the source separation in metres and \varepsilon the dimensionless relative permittivity of the medium; in vacuum, \varepsilon=1. The simple formula below assumes a uniform medium throughout the space considered. An arbitrary boundary between materials cannot be replaced by one number without analysis.

Coulomb's law gives force magnitude: F=k|q_1q_2|/(\varepsilon r^2), where k\approx9\cdot10^9\,\text{Н}\cdot\text{м}^2/\text{Кл}^2. Charges in C and distance in m give force in N. Doubling one charge doubles F; doubling the distance reduces it by a factor of four. A magnitude cannot be negative: use charge signs to determine direction separately. The limit r → 0 warns of the point model's limits for real objects of finite size.

Another useful constant is the vacuum permittivity \varepsilon_0\approx8{,}85\cdot10^{-12}\,\text{Ф/м}: k=1/(4\pi\varepsilon_0). This expresses the same interaction coefficient differently, not an additional effect of the medium. The farad is a unit of capacitance, introduced in the next topic. For now, distinguish \varepsilon_0, with units F/m, from relative \varepsilon, a dimensionless factor.

When can we simply divide the vacuum force by ε? The dielectric must be uniform, isotropic—the same in every direction—and linear, with polarisation proportional to the field. Near a material boundary or conductor, induced charges have a different distribution. Imagine a metal wall near the original pair: it changes the field and its directions even though the original charge separation is unchanged. A single replacement of ε is no longer enough.

Place a small test charge at the chosen point and measure the force. Doubling the test charge doubles the force, so their ratio remains a property of the sources. This is electric field strength: \vec E=\vec F/q, in N/C. E points in the force direction for a positive test charge. The test charge must be small enough to barely disturb the sources or their charge distribution; otherwise, we would measure a different field.

The definition gives \vec F=q\vec E. Positive q preserves the direction; negative q reverses it. An electron therefore feels a force opposite to E, although the field does not 'turn around' when it appears. With the sources unchanged, doubling the test charge doubles the force while E stays fixed. Removing the test charge removes the force on that object, but the source field remains. This separates a field property from its interaction with a particular test object.

Call the source charge Q to distinguish it from the test charge q. Dividing the Coulomb force by the magnitude of the test charge gives E=k|Q|/(\varepsilon r^2). The q cancels, as it should for a property of the source. The field points away from positive Q and towards negative Q. At twice the distance, its magnitude is four times smaller; doubling |Q| doubles it. The point-source and uniform linear-medium assumptions are the same as for the force.

Electric field strength can also be measured in V/m: 1\,\text{Н/Кл}=1\,\text{В/м}. It is the same quantity expressed in two ways. N/C is convenient when force on a test charge is known; V/m will be convenient when potential difference and distance along a uniform field are known. The new unit does not add a new law. In the next lesson, the volt as a joule per coulomb connects energy and force descriptions.

Why does a dielectric change the field? Its positive and negative charges shift slightly relative to each other, or bound pairs of charges align. This is polarisation; it does not require charges to flow freely across the whole sample as in a metal. The polarised material creates its own field, which adds to the original field. Let E₀ be the field without the dielectric and E the field after it fills the space. Comparing them requires the same specified sources.

With free charges and geometry unchanged, completely filling the space uniformly with a linear isotropic dielectric gives \varepsilon=E_0/E. Doubling ε means halving E under these conditions. But a power supply holding voltage constant can change the free charges, breaking the comparison. The formula is not universal for every object immersed in a dielectric; first establish what stays fixed and where the material boundaries are.

How do we find the field from several sources? At the same point, draw \vec E_1,\vec E_2,\ldots separately, then apply superposition: \vec E=\vec E_1+\vec E_2+\ldots. These are vectors, so two nonzero contributions can sum to zero. Adding positive magnitudes is valid only when directions agree; opposite fields partly cancel. For angled sources, use a vector diagram or components instead of guessing a sign from numbers.

Place two equal positive charges symmetrically on either side of a point. At the midpoint, the fields have equal magnitudes and opposite directions, so E = 0. No direction could be selected without breaking symmetry. Now replace one source with a negative charge: between the charges, both fields point from positive to negative and their magnitudes add. Changing one sign turns cancellation into reinforcement even though distances and charge magnitudes are unchanged.

A positive test charge of 2 nC feels a force of 6 μN to the right. Find the electric field strength.

The charge is positive, so E points with the force: to the right.

Convert units: q=2·10⁻⁹ C, F=6·10⁻⁶ N.

By definition, E=F/q=3000 N/C.

Check: twice the charge would feel twice the force, but F/q would stay the same.

E=3000 N/C to the right.

An isolated system's total charge is conserved, although individual objects can exchange electrons.

E describes the source field; F = qE also accounts for the sign and magnitude of the test charge.

Coulomb's law and the field reduction described by ε require a specific medium model. Fields from multiple sources add as vectors.

Next, we explore how a field transfers energy as a charge moves.