Question

Two points charges $+9e$ and $+e$ are kept $16cm$ apart to each other. Where a third charge $q$ is placed between them so that the system is in the equilibrium state-

• A. $24cm$ form $+9e$
• B. $12cm$ form $+9e$
• C. $24cm$ form $+e$
• D. $12cm$ form $+e$

Answer 1

Answer: (D)

$12cm$ form $+e$

For equilibrium state

$\frac{9\times q_c}{x^2}=\frac{q\times e}{{(16-x)}^2}$

or, $\frac{9}{x^2}=\frac{1}{{(16-x)}^2}$

or, $\frac{3}{x}=\frac{1}{16-x}$

or, $3(16-x)=x$

or, $48-3x=x$

or, $4x=48$

or, $x=12$ (Ans)

Now, from $x$ the distance will be $(16-12)cm=4cm$.