The three fixed charges are related by rotations through

2 π /3

about the origin. Their three Coulomb's law forces on a charge placed at the origin have equal magnitudes and directions separated by

2 π /3

, so their vector sum vanishes. Thus

r = 0

ii

Solution

Let

ρ = a / 3

be the distance from the origin to each fixed charge, and denote their position vectors by

R_{j}

. For a small planar displacement

r

\frac{1}{∣ R _{j} - r ∣} = \frac{1}{ρ} + \frac{R _{j} \cdot r}{ρ ^{3}} + \frac{3 ( R _{j} \cdot r ) ^{2} - ρ ^{2} ∣ r ∣ ^{2}}{2 ρ ^{5}} + O (∣ r ∣^{3})

can be summed using the threefold symmetry relations

\sum_{j} R_{j} = 0, \sum_{j} R_{j} R_{j}^{T} = \frac{3 ρ ^{2}}{2} I .

The electric potential due to the fixed charges is consequently

Φ (r) = \frac{q}{4 π ϵ _{0}} (\frac{3}{ρ} + \frac{3∣ r ∣ ^{2}}{4 ρ ^{3}} + O (∣ r ∣^{3})) .

U = q Φ

. Because the quadratic coefficient is positive, the origin is a strict local minimum of

U

, and the equilibrium is stable.

Taking the gradient of the quadratic potential energy found in part (ii) resolves the linearized force into identical

x

and

y

components:

F_{x} = - \frac{3 q ^{2}}{8 π ϵ _{0} ρ ^{3}} x, F_{y} = - \frac{3 q ^{2}}{8 π ϵ _{0} ρ ^{3}} y .

Each component therefore performs simple harmonic motion with

ω^{2} = \frac{3 q ^{2}}{8 π ϵ _{0} m ρ ^{3}} = \frac{9 3}{8 π} \frac{q ^{2}}{m ϵ _{0} a ^{3}} .

ω = \frac{3 ^{5/4}}{8 π} \frac{∣ q ∣}{m ϵ _{0} a ^{3}}, A = \frac{3 ^{5/4}}{8 π} .

There are no discussions about this article yet.

There are currently no matching articles.