P22_010.PDF

10. There is no equilibrium position for q 3 between the two ﬁxed charges, because it is being pulled by one

and pushed by the other (since q 1 and q 2 have diﬀerent signs); in this region this means the two force

arrows on q 3 are in the same direction and cannot cancel. It should also be clear that oﬀ-axis (with the

axis deﬁned as that which passes through the two ﬁxed charges)there are no equilibrium positions. On

the semi-inﬁnite region of the axis which is nearest q 2 and furthest from q 1 an equilibrium position for

q 3 cannot be found because

q 1 |

q 2 |

k |

q 1 q 3 |

x 2

−

k |

q 2 q 3 |

( d + x ) 2

with d =10cmand x assumed positive. We set this equal to zero, as required by the problem, and

cancel k and q 3 . Thus, we obtain

q 1 |

x 2 −

q 2 |

( d + x ) 2 =0 =

d + x

q 2

q 1

⇒

which yields (after taking the square root)

d + x

= √ 3=

⇒

x =

√ 3 − 1 ≈

14 cm

for the distance between q 3 and q 1 ,so x + d (the distance between q 2 and q 3 )is approximately 24 cm.

and the magnitude of force exerted by q 2 is everywhere (in that

region)stronger than that exerted by q 1 on q 3 . Thus, we must look in the semi-inﬁnite region of the

axis which is nearest q 1 and furthest from q 2 , where the net force on q 3 has magnitude