question_answer

Three charges +Q, q, +Q are placed respectively, at distance, 0, \[d/2\] and d from the origin, on the x-axis. If the net force experienced \[by+Q\], placed at \[x=0\], is zero then value of q is: [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

A) +Q/2           

B) -Q/4

C) +Q/4           

D) -Q/2

Correct Answer: B

Solution :

\[\frac{kQq}{{{\left( \frac{d}{2} \right)}^{2}}}\,\,=\,\,\frac{k{{Q}^{2}}}{{{d}^{2}}}\] \[q\,\,=\,\,\frac{Q}{4}\,\,\frac{{{d}^{2}}}{{{d}^{2}}}\] \[q\,\,=\,-\,\frac{Q}{4}\,\,\] Option (2) is correct.