Charges 4Q, q and Q are placed along \[x\]-axis at positions \[=0,x=1/2\] 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 :
Total force acting on charge Q \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{{{(l/2)}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4Q.Q}{{{(l)}^{2}}}\] According to question \[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{qQ}{{{(l/2)}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{4Q.Q}{{{(l)}^{2}}}=0\] \[\frac{4q}{{{l}^{2}}}+\frac{4Q}{{{l}^{2}}}=0\] \[\therefore \] \[q=-Q\]