 Solution

ID: past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013-resit/7/ii/solution

With

q_{2} = - q_{1}

, both field contributions between the charges point the same way. Their signed force on

q_{3}

F_{x} (f) = \frac{k _{e} q _{1} q _{3}}{L ^{2}} [\frac{1}{( 1 + f ) ^{2}} + \frac{1}{( 1 - f ) ^{2}}] = \frac{2 k _{e} q _{1} q _{3}}{L ^{2}} \frac{1 + f ^{2}}{( 1 - f ^{2} ) ^{2}}, k_{e} = \frac{1}{4 π ϵ _{0}} .

Since

f

is dimensionless, the small-displacement expansion is

∣ f ∣ ≪ 1

F_{x} (f) = \frac{2 k _{e} q _{1} q _{3}}{L ^{2}} (1 + 3 f^{2} + O (f^{4})) .

The force at the midpoint is nonzero and the first variation is quadratic, not linear. In particular, the midpoint is not an equilibrium supporting a linear restoring-force approximation. The printed comparison of

f

with

L

is dimensionally inconsistent; the expansion above keeps their distinct roles.

 New to topics? Read the docs here!