Question

Two particles A and B with charges q and 2q, respectively, are placed on a smooth table with a separation d. A third particle C is to be clamped on the table in such a way that the  particles A and B remain at rest on the table under electrical forces. What should be the charge on C and where should it be clamped? 

Answer 1

For equilibrium,

\[\vec{F}_{AC}  +  \vec{F}_{CB}  = 0\]

Let the charge at point c be θ.

 

\[\frac{q\theta}{4\pi  \in_0 x^2} + \frac{2q\theta}{4\pi  \in_0 \left( d - x \right)^2} = 0\]

 

\[\text{Again },    \vec{F}_{AC}  =  \vec{F}_{CB} \] 

\[\text{So} ,   \frac{1}{x^2} = \frac{2}{\left( d - x \right)^2}\]

\[\text{Or 2 x}^2  =  \left( d - x \right)^2 \] 

\[ \text{Or  }\sqrt{2}x = d - x\] 

\[ \text{Or x } = (\sqrt{2} - 1)d\]

For a charge at rest,

\[\vec{F}_{AC}  =  \vec{F}_{CB} \] 

\[\frac{1}{4\pi  \in_0}\frac{q\theta}{[(\sqrt{2} - 1)d ]^2} + \frac{1}{4\pi  \in_0}\frac{q \times 2q}{d^2}\] 

\[ = 0\] 

\[ \text{ Or } \theta = (6 - 4\sqrt{2})q\]