Capacitors

Transfer electrons from one metal plate to another. The pair's total charge remains zero, but a field and voltage appear between the plates. Why does separating charge require work, and what determines the energy stored?

Two separated conducting plates form a capacitor. If one carries +q and the other −q, their total charge is zero, but the charges occupy different positions. Their fields therefore need not cancel between the plates. Separating more opposite charge against its attraction requires work, which increases electric field energy. Thus zero net charge is compatible with stored energy: the spatial arrangement matters.

How can we compare two devices: which separates more charge at the same voltage? The ratio C=q/U is called capacitance. Here q is the magnitude of charge on one plate in C, not the pair's zero total; U is voltage magnitude in V. The unit is the farad, 1\,\text{Ф}=1\,\text{Кл/В}. A larger C means more charge at the same voltage, or less voltage at the same charge. Keep the device and medium properties unchanged when making this comparison.

For a linear capacitor with unchanged geometry, voltage increases in proportion to separated charge. Thus q=CU with constant C: doubling U doubles q. If charge and voltage increase together, their ratio stays the same; capacitance has not increased. It describes the same proportional relationship, not the current amount of charge. At zero voltage an ideal uncharged capacitor has q = 0, but its capacitance remains a property of the device.

Why can wider plates store more charge at a given voltage? A larger area can hold more distributed charge at the same field strength; use the overlapping area S in m². Increasing the gap d at fixed charge increases potential difference and therefore reduces C. A dielectric between the plates polarizes and also changes the relation between q and U; its relative permittivity is ε. Geometry and the medium explain capacitance even before we use the formula.

For a parallel-plate capacitor, C=\varepsilon_0\varepsilon S/d, with \varepsilon_0 in F/m. Doubling S or ε doubles capacitance; doubling d halves it. The units F/m·m²/m give F. Assume a uniform linear dielectric fills the entire gap, d is small compared with the plate dimensions, and edge effects are small. The limit d → 0 cannot be realized literally: contact or electrical breakdown destroys the model of isolated conductors.

During charging, voltage rises linearly from zero to U, so transferring each further portion of charge requires more work. The average voltage during this charging is U/2, giving field energy W=qU/2=CU^2/2=q^2/(2C) in joules. Doubling U at fixed C multiplies W by four. The three forms are equal because q = CU. If capacitance itself changes, however, the conclusion depends on what remains fixed: q or U.

Charge the capacitor and disconnect the source. If the plates are isolated and there is no leakage, q remains constant. Increasing C then reduces energy W=q^2/(2C) and voltage q/C. For example, bringing the plates closer reduces the voltage needed for that charge separation. The energy has not vanished: the field can do mechanical work. If external forces change the geometry, include their work in the overall energy balance.

Leave the capacitor connected to an ideal source that holds U fixed. Now increasing C draws in charge q = CU, and energy W=CU^2/2 increases. Doubling C at the same U doubles the energy, opposite to the isolated-plate case. There is no contradiction: the system boundary has changed, and the source transfers charge and energy. To calculate mechanical work when the gap changes, tracking W alone is insufficient; the source also participates.

Connect initially uncharged capacitors in series. The intermediate node is isolated from external branches and remains neutral: as many electrons leave one of its plates as arrive on the other. Those plates therefore carry equal opposite charges, and the capacitors have equal charge magnitudes q. Voltages across the series sections add. If the intermediate node is charged beforehand, this simple equality of magnitudes must be reconsidered.

For two series components, q_1=q_2=q and U=U_1+U_2. Replace each voltage with q/C: q/C=q/C_1+q/C_2, giving 1/C=1/C_1+1/C_2. We started by adding voltages for the same charge, so reciprocal capacitances arise naturally. At the same q, the smaller C has the larger voltage. Two identical components have half the capacitance of one: the same charge requires twice the total U.

For two components, rewrite the result as C=C_1C_2/(C_1+C_2). The total capacitance is smaller than either individual value, a useful arithmetic check. If one capacitance is much larger, nearly all the voltage is across the smaller one, and the total approaches that smaller capacitance from below. A very large C in a series chain therefore does not make the total very large: the smaller capacitor causes the greater voltage rise.

Now connect both pairs of plates to the same pair of nodes. This is a parallel connection: U=U_1=U_2, because the same two potentials are being compared. The total charge on the positive plates is q=q_1+q_2. Different C values at one U give different charges: the larger capacitance takes more charge. This is not the total charge of all positive and negative plates together, which can remain zero.

Substitute q_1=C_1U and q_2=C_2U into the charge sum: q=(C_1+C_2)U, so C=C_1+C_2. An added branch increases charge at the same voltage, like extra plate area. The total C exceeds either individual value and doubles for two identical components. Compare this with series connection: there, voltages add at one charge; here, charges add at one voltage.

Capacitors of 2 μF and 3 μF are connected in parallel to 10 V. Find the total capacitance and total charge on the positive plates.

In parallel, both capacitors have the same voltage.

Capacitances add: C=2+3=5 μF.

q=CU=5·10⁻⁶×10=50·10⁻⁶ Кл=50 мкКл.

Check each branch: 20 μC + 30 μC = 50 μC.

C=5 мкФ, q=50 мкКл.

Capacitance relates charge on one plate to voltage. Zero total charge on the pair does not rule out a field or energy.

Analyse energy changes together with the conditions: isolation holds q fixed; an ideal source holds U fixed.

Series connection adds voltages at equal charges; parallel connection adds charges at a common voltage.

Next, consider continuous charge transport: electric current.