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BPO 2025, Теория — Задача 3. Charged spheres and springs

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BPO7 Problems 26-30 June 2025, Ohrid · 26 юни 2025 г. · 25 т.

Условие

Two identical small spheres, each of mass $m = 1.00\ \mathrm{g}$ , are electrically charged with the same unknown charge $q$ , and are suspended from the same point by two identical, inextensible wires of length $L_0 = 0.200\ \mathrm{m}$ . Each wire forms an angle of $\alpha = 30^{\circ}$ with the vertical. Wires are non-conducting.

a) Find the mathematical expression of the electrostatic force acting on each sphere, as well as its numerical value. [2 т.]

b) Find the mathematical expression of the electric charge on each sphere, in terms of the known quantities, as well as its numerical value. [2,5 т.]

c) Find the mathematical expression of the energy stored in the system, assuming the gravitational potential energy is zero at a distance $L_0$ below the point from which the wires are suspended. [3,5 т.]

Now consider 5 identical spheres, each with the same properties and electric charge as above. Four of them are fixed on a horizontal surface, at the corners of a square of side $L_0$ . The fifth sphere is attached to the other four by four identical springs, of negligible mass, natural (unstretched) length $L_0$ , and unknown spring constant $k$ . The springs remain straight (they do not bend). The system forms a regular square pyramid. When the system is in equilibrium, each spring is compressed by a distance $\Delta L$ . The springs are non-conducting.

d) Find the mathematical expression of the spring constant $k$ , in terms of $m, q, L_0, \Delta L$ , and known constants. [5 т.]

e) Find the mathematical expression of the electrostatic energy of the system in terms of $q, L_0, \Delta L$ and known constants. [2 т.]

f) The sphere at the top of the pyramid is displaced vertically by a very small distance $h$ , where $h \ll L_0$ . Show that the restoring force acting on the sphere immediately after it is released is of the form $F_r = C\cdot h$ where $C$ is a constant. Find the mathematical expression for the constant $C$ . [10 т.]

Note: The electric permittivity of the medium $\varepsilon_0 = 8.854\cdot10^{-12}\ \mathrm{F/m}$ and the gravitational acceleration $g = 9.80\ \mathrm{ms^{-2}}$ are known.

Отговори

Решение

Solution:

$\frac{F_e}{mg}=\tan\alpha$ (1.5)

$F_e=mg\cdot\tan\alpha$ (0.2)

$F_e=5{,}66\cdot10^{-3}\ \mathrm{N}.$ (0.3)

$F_e=\frac{q^2}{4\pi\varepsilon_0d^2}$ (1.0)

$d^2=4L_0^2\sin^2\alpha$ (0.7)

$q=\pm4L_0\sin\alpha\cdot\sqrt{\pi\varepsilon_0F_e}$ (0.5)

$q=\pm159\ \mathrm{nC}$ (0.3)

$W_{pg}=2mgL_0(1-\cos\alpha)$ (1.5)

$W_e=\frac{q^2}{8\pi\varepsilon_0L_0\sin\alpha}$ (1.5)

$W=W_{pg}+W_e=2mgL_0(1-\cos\alpha)+\frac{q^2}{8\pi\varepsilon_0L_0\sin\alpha}$ (0.5)

(1.5 for the drawing)

$F_e=\frac{q^2}{4\pi\varepsilon_0L^2}$ (0.5)

Vertical cross-section of the pyramid: the top sphere joined by springs to two base spheres; forces F_e, F_el along the spring and mg downward at the top sphere; spring length L=L_0-ΔL, angle β at the base, base diagonal √2 L_0 dashed.

$F_{el}=k\cdot\Delta L$ (0.5)

$mg=4\left(\frac{q^2}{4\pi\varepsilon_0L^2}+k\cdot\Delta L\right)\sin\beta$ (1.0)

$mg=4\left(\frac{q^2}{4\pi\varepsilon_0(L_0-\Delta L)^2}+k\cdot\Delta L\right)\sin\beta$ (0.5)

mg=4\left(\frac{q^2}{4\pi\varepsilon_0(L_0-\Delta L)^2}+k\cdot\Delta L\right)\sqrt{1-\frac{L_0^2}{2L^2}}

k\cdot\Delta L=\frac{mg}{4}\cdot\left(1-\frac{L_0^2}{2L^2}\right)^{-\frac{1}{2}}-\frac{q^2}{4\pi\varepsilon_0(L_0-\Delta L)^2}

$k=\frac{mg}{4\cdot\Delta L}\cdot\left(1-\frac{L_0^2}{2(L_0-\Delta L)^2}\right)^{-\frac{1}{2}}-\frac{q^2}{4\pi\varepsilon_0\cdot\Delta L(L_0-\Delta L)^2}$ (1.0)

$W_e=\frac{q^2}{4\pi\varepsilon_0}\left[\frac{4}{L_0-\Delta L}+\frac{4}{L_0}+\frac{2}{\sqrt{2}L_0}\right]$ (1.5)

$W_e=\frac{q^2}{2\pi\varepsilon_0}\left[\frac{2}{L_0-\Delta L}+\frac{2}{L_0}+\frac{1}{\sqrt{2}L_0}\right]$ (0.5)

If the ball at the vertex descends on a very small distance $h$ , the length of the springs decreases with $h\sin\beta$ . (0.5)

The supplementary vertical component of the elastic is:

$F_1=4k\sin^2\beta\cdot h$ (1.0)

The new electric force will be:

$F_e'=\frac{q^2}{4\pi\varepsilon_0(L-h\sin\beta)^2}$ (1.0)

But

$\frac{1}{L-h\sin\beta}=\frac{L+h\sin\beta}{L^2-(h\sin\beta)^2}\cong\frac{L+h\sin\beta}{L^2}$ (1.0)

$\frac{1}{(L-h\sin\beta)^2}\cong\frac{(L+h\sin\beta)^2}{L^4}=\frac{L^2+2Lh\sin\beta+(h\sin\beta)^2}{L^4}\cong\frac{L^2+2Lh\sin\beta}{L^4}$ (1.0)

so:

$F_e'\cong\frac{q^2}{4\pi\varepsilon_0}\cdot\frac{L^2+2Lh\sin\beta}{L^4}$ (0.5)

The supplementary vertical component of the electric force is:

$F_2=4(F_e'-F_e)\sin\beta$ (1.0)

F_2\cong4\left(\frac{q^2}{4\pi\varepsilon_0}\cdot\frac{L^2+2Lh\sin\beta}{L^4}-\frac{q^2}{4\pi\varepsilon_0L^2}\right)\sin\beta

F_2\cong4\frac{q^2}{4\pi\varepsilon_0L^2}\left(\frac{L^2+2Lh\sin\beta}{L^2}-1\right)\sin\beta

F_2\cong4\frac{q^2}{4\pi\varepsilon_0L^2}\cdot\frac{2h\sin^2\beta}{L}

$F_2\cong\frac{2q^2}{\pi\varepsilon_0}\cdot\frac{\sin^2\beta}{L^3}\cdot h$ (1.5)

The restoring force is:

$F_r=F_1+F_2\cong4k\sin^2\beta\cdot h+4\frac{2q^2}{4\pi\varepsilon_0}\cdot\frac{\sin^2\beta}{L^3}\cdot h$ (0.5)

$F_r\cong4\sin^2\beta\left(k+\frac{q^2}{2\pi\varepsilon_0L^3}\right)\cdot h$ (1.0)

F_r=C\cdot h,\ \text{with}\ C\cong4\sin^2\beta\left(k+\frac{q^2}{2\pi\varepsilon_0L^3}\right)

$\sin^2\beta=\frac{L^2-\frac{L_0^2}{2}}{L^2}=1-\frac{L_0^2}{2(L_0-\Delta L)^2}$ (0.5)

$C\cong4\left(1-\frac{L_0^2}{2(L_0-\Delta L)^2}\right)\left(k+\frac{q^2}{2\pi\varepsilon_0L^3}\right)$ (0.5)

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf