question_answer

Three charges \[4q\], \[Q\] and q are placed in a straight line of length \[l\] at points distant 0, 1/2 and \[l\]respectively. What should be \[Q\] in order to make the net force on q equal to zero?

A)  \[-4q\]                

B)         \[-2q\]

C)  \[-\frac{q}{2}\]                

D)         \[-q\]  

Correct Answer: D

Solution :

  net\[F=0\] \[\therefore \]  \[0=\frac{4q\times q}{{{l}^{2}}}+\frac{Q\times q}{{{(l/2)}^{2}}}\]                    \[=\frac{4{{q}^{2}}}{{{l}^{2}}}+\frac{4Qq}{{{l}^{2}}}=\frac{4{{q}^{2}}+4Qq}{{{l}^{2}}}\] \[\therefore \]        \[Q=-q\]