Question

Three small positive electrical charges A, B and C are placed along a straight line. The distance between A and B is 12 cm. A and B have equal charges while C has twice the charge of each of A and B.

At what distance from B should C be placed in order for the resulting force on B to be zero?

Answer 1

Solution:

A and B have charge q, C has charge 2q, correct? Assume that is correct.

A's field at B points right. C's field at B points left so the fields subtract.

$\dfrac{q}{0.12^2}-\dfrac{2q}{r^2}=0$ where r is the distance from C to B

$q*r^2=2q*0.12^2$

$r^2=2*0.12^2 =0.288$

r = $\sqrt{\smash[b]{0.288}}$ = 0.17m  = 17cm from B to C