Question

q1=+10 nC and q2=-5 nC are separated by 0.10 m dinstance. If a third charge q3=2 nC is placed along with the first two charges, where should it be so that the net electric force of q1 and q2 on q3 is zero (0)?

Answer 1

$F_{23}=\dfrac{kq_2q_3}{r^{2}}$

$F_{13}=\dfrac{kq_1q_3}{r^{2}}$

let distance between $q_2\ and \ q_3\ be \ x.$

so the force of the charge by $q_2$ on$\ q_3$ is denoted by $F_{23}$ and the force on charge $q_1$ on $q_3$ is denoted by $F_{13}$ .

Distance between $q_3\ and \ q_1$ is given $0.10+x$

Distance between $q_1\ and \ q_2$ is given by 0.10 meters

now equation both the forces we get -

$F_{13}=F_{23}$

$\dfrac{kq_1q_3}{(0.10+x)^{2}}=\dfrac{kq_2q_3}{x^{2}}$

$=$ $\sqrt{0.5}$ $=\dfrac{x}{x+0.10}$

$=0.70=\dfrac{x}{x+0.10}$

=$0.70x+0.70\times 0.10=x$

$=0.70\times 0.10=x-0.70x$

$=0.07=0.30x$

$=x=0.23m$