 Solution

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

Put

q_{1}

x = - L

q_{2}

x = + L

, and

q_{3}

x = f L

, with

- 1 < f < 1

. For a nonzero test charge, the force vanishes when the electric field vanishes:

\frac{q _{1}}{( 1 + f ) ^{2}} = \frac{q _{2}}{( 1 - f ) ^{2}} .

For equal-sign nonzero

q_{1}, q_{2}

, take the positive square root

s = q_{2} / q_{1}

. The distances are positive, so

1 - f = s (1 + f)

and

f = \frac{1 - s}{1 + s} .

For

s > 1

the zero moves toward the weaker charge

q_{1}

; equal charges give

f = 0

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