Question

Three charged particles, $A$, $B$, and $C$, are kept apart by distance d from each other as shown in figure. If particle $B$ is attracted to particle $A$, what can we predict about the charge, $q_A$, of particle $A$?

• A. $q_A<-q$
• B. $-q<q_A<0$
• C. $q_A=0$
• D. $0<q_A<+q$
• E. $q_A>+q$

Answer 1

The correct option is E $q_A>+q$

Particle at $B$ is going towards $A$, it means a larger electrostatic force of attraction is exerted on it by particle at $A$ than by particle at $B$ . Electrostatic force between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by

$E=\frac{1}{4\pi \epsilon }\frac{q_1{q}_2}{r^2}$

i.e. $E$ depends upon medium between charges ,distance between them and magnitude of charges,

as for two charges at points $A$ and at $C$ , medium and distance from $B$ are same, therefore larger force will be exerted by greater charge and as given $-q$ is attracted towards $A$ so $q_A>+q$.