Question

Three particles, A , B , and C , are set in a line, with a distance of d between each of them, as shown above. If particle B is attracted more towards particle A, what can we say 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 $q_A>+q$

Given that negative particle B is attracted to particle A , from this it is clear that particle A is a positive charge .

Now by Coulomb's law , electrostatic force between particle A and B ,

$F=k\frac{q_{A}q}{d^2}$

and force between particle B and C ,

$F^{\prime }=k\frac{qq}{d^2}$

as the particle B is attracted towards A , therefore A exerts a greater force on B than C i.e. $F>F^{\prime }$ ,

or $k\frac{q_{A}q}{d^2}$> $k\frac{qq}{d^2}$

or $q_A>+q$