Question

Two force point charges $+q$ and $+4q$ are a distance $\ell$ apart. A third charge $q_1$ is so placed that the entire system is in equilibrium. Find the location, magnitude and sign of the third charge.

• A. $q_1=\frac{4}{9}q$
• B. $q_1=\frac{-4}{9}q$
• C. $q_1$ is to be placed at a distance $\ell /3$ from $q$.
• D. $q_1$ is to be placed at a distance $\frac{2\ell }{3}$ from $4q$.

Answer 1

Answer: (B), (C), (D)

The correct options are
B $q_1=\frac{-4}{9}q$
C $q_1$ is to be placed at a distance $\ell /3$ from $q$.
D $q_1$ is to be placed at a distance $\frac{2\ell }{3}$ from $4q$.

To get zero net force on placed charge it must be negative.

On balancing the forces,

$\frac{Kq{q}^{\prime }}{(l-x{)}^2}=\frac{K{q}^{\prime }\left(4q\right)}{x^2}$

$x^2=4(l-x{)}^2$

$x^2=4l^2+4x^2-8lx$

$3x^2-8lx+4l^2=0$

$x=\frac{8l\pm \sqrt{64l^2-48l^2}}{6}=\frac{8l\pm 4l}{6}=2l$ and $\frac{2l}{3}$

$x=\frac{l}{3}$ is the suitable answer.

Now, on balancing the forces on charge 4q

$\frac{K\left(q^{\prime }\right)4q}{{\left(\frac{2l}{3}\right)}^2}=\frac{K\left(q\right)\left(4q\right)}{l^2}$

$q^{\prime }=-\frac{4q}{9}$