question_answer

Two equal point changes,\[Q=+\sqrt{2}\mu C\]are placed at each of the two opposite corners of a square and equal point charges q at each of the other two comers. The value of q, so that the resultant force on Q is zero is:

A)  \[+0.5\mu C\]                  

B)  \[-0.5\mu C\]   

C)         \[1\mu C\]                        

D)         \[-1\mu C\]

E)  none of above

Correct Answer: B

Solution :

From Coulombs law \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\,\frac{{{Q}^{2}}}{2{{a}^{2}}}\] \[{{F}_{1}}\]and\[{{F}_{2}}\]will be directed as shown, for this both q should be negative \[{{F}_{1}}={{F}_{2}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{Qq}{{{a}^{2}}}\] \[{{F}_{12}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{\sqrt{2}Qq}{{{a}^{2}}}\] For equilibrium of Q, we have \[F=-{{F}_{12}}\] \[\frac{{{Q}^{2}}}{2{{a}^{2}}}=\frac{\sqrt{2}Qq}{{{a}^{2}}}\] \[\Rightarrow \]               \[q=\frac{Q}{2\sqrt{2}}\] Given,       \[Q=+\sqrt{2}\mu C\] \[\therefore \]  \[q=-\frac{\sqrt{2}\mu C}{2\sqrt{2}}=-0.5\,\mu C\]