Two free positive charges $4q$ and $q$ are a distance $l$ apart. What charge $Q$ is needed to achieve equilibrium for the entire system and where should it be placed form charge $q$ ?
$Q = \frac{4}{9} q\,(negative)\, at \frac{l}{3}$ distance
$Q = \frac{4}{9} q\,(positive)\, at \frac{l}{3}$ distance
$Q = q\,(positive)\, at \frac{l}{3}$ distance
$Q = q\,(negative)\, at \frac{l}{3}$ distance
Three charges $+Q, q, +Q$ are placed respectively, at distance, $0, \frac d2$ and $d$ from the origin, on the $x-$ axis. If the net force experienced by $+Q$, placed at $x = 0$, is zero, then value of $q$ is
Three point charges are placed at the corners of an equilateral triangle. Assuming only electrostatic forces are acting
There are two charges $+1$ microcoulombs and $+5$ microcoulombs. The ratio of the forces acting on them will be
$ABC$ is a right angled triangle in which $AB = 3\,cm$ and $BC = 4\,cm$. And $\angle ABC = \pi /2$. The three charges $ + 15,\; + 12$ and $ - 20\,e.s.u.$ are placed respectively on $A$, $B$ and $C$. The force acting on $B$ is.......$dynes$
Three charges are placed at the vertices of an equilateral triangle of side ‘$a$’ as shown in the following figure. The force experienced by the charge placed at the vertex $A$ in a direction normal to $BC$ is