BPO 2026, Теория — Задача 4. Spring circuit

• A) $Q' = 2C\Delta V$
• B) $F = \frac{Q^2}{2\varepsilon_0 A}$
• C) $k = \frac{8C(\Delta V)^2}{d^2}$
• D) $k = \frac{8C\Delta V^2}{d^2}$

A) Capacitance

$C = \varepsilon_0 A / d$

When the distance becomes d/2:

$C' = \varepsilon_0 A / (d/2) = 2C$ 1p

$Q' = C'\,\Delta V = 2C\,\Delta V$ 1p

B) Force between the plates

$U = Q^2 / (2C)$

For d:

$U_1 = Q^2 d / (2\,\varepsilon_0 A)$ 1p

For d/2:

$U_2 = Q^2 (d/2) / (2\,\varepsilon_0 A)$ 1p

$\Delta U = -\,Q^2 d / (4\,\varepsilon_0 A)$ 2p

$W = F\,d/2$ and $W = -\Delta U$ 6p

$F = Q^2 / (2\,\varepsilon_0 A)$ 2p

C) Force and springs

Displacement:

$d/2 = F/k + F/k = 2F/k \rightarrow F = kd / 4$

Electric force:

$F = Q^2 / (2\,\varepsilon_0 A)$

$Q = C'\,\Delta V$

$C' = 2C$

$F = C'^2\Delta V^2 / (2\,\varepsilon_0 A)$

$F = (2C)^2\,\Delta V^2 / (2\,\varepsilon_0 A)$

Since $\varepsilon_0 A = Cd$:

$F = 2C\,(\Delta V)^2 / d$ 5p

Equation:

$kd / 4 = 2C\,(\Delta V)^2 / d \rightarrow k = 8C\,(\Delta V)^2 / d^2$ 1p

D. Energy

Initially:

$U = 1/2\ C\Delta V^2$

Finally:

$U' = 1/2 \bullet 2C \bullet \Delta V^2 = C\Delta V^2$

$\Delta U = 1/2\ C\Delta V^2$ 1p

Spring energy:

$U_\mathrm{spring} = 2 \bullet (1/2)\ k\,(d/4)^2 = kd^2 / 16$ 2p

Equating:

$1/2\ C\Delta V^2 = kd^2 / 16 \rightarrow k = 8C\Delta V^2 / d^2$ 2p