Question 1165901
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The given charge q is at (0,0); the charge 9q is at (1,0).<br>
Let the added charge q be at (b,0).<br>
The distance from (0,0) to (b,0) is b; the distance from (1,0) to (b,0) is 1-b.<br>
The force between any two charges is directly proportional to the sizes of both charges and inversely proportional to the square of the distance between them.<br>
{{{(q)(q)/(b^2) = (q)(9q)/(1-b)^2}}}
{{{q^2/b^2 = 9q^2/(1-b)^2}}}
{{{9q^2/q^2 = (1-b)^2/b^2}}}
{{{9 = (1-b)^2/b^2}}}
{{{3 = (1-b)/b}}}
{{{3b = 1-b}}}
{{{4b = 1}}}
{{{b = 1/4}}}<br>
ANSWER: The additional charge q should be placed at (1/4,0) to make the net force on it zero.<br>