question_answer

Charges 4 Q, q and Q are placed along x-axis at position \[x=o,x=l/2\] and\[x=l,\] respectively. Find the value of q, so that force on charge Q is zero.

A)  \[Q\]               

B)  \[\frac{Q}{2}\]

C)  \[-\frac{Q}{2}\]             

D)  \[-Q\]

Correct Answer: D

Solution :

 From Coulombs law, the force acting between two charges \[({{q}_{1}}\,,\,{{q}_{2}})\] separated at a distance r is given by \[F=\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}\frac{{{q}_{1}}\,\,{{q}_{2}}}{{{r}^{2}}}\] Total force acting on charge Q is \[F=\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}\frac{q\,Q}{{{(l/2)}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4\,Q\,\,.\,\,Q}{{{(l)}^{2}}}\] According to question, F = 0 \[\therefore \] \[\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}\frac{qQ}{{{(l/2)}^{2}}}+\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}.\frac{4\,{{Q}^{2}}}{{{(l)}^{2}}}=0\] \[\Rightarrow \] \[\frac{4\,q}{{{l}^{2}}}+\frac{4\,Q}{{{l}^{2}}}=0\]