A) \[Q/2\]
B) \[-Q/2\]
C) \[Q/4\]
D) \[-Q/4\]
Correct Answer: D
Solution :
[d] Let charge q be placed at mid-point of line AB as shown below. |
Also \[AB=x\] (say) |
\[\therefore \] \[AC=\frac{x}{2},BC=\frac{x}{2}\] |
For the system to be in equilibrium, |
\[{{F}_{Qq}}+{{F}_{QQ}}=0\] |
\[\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{Qq}{{{(x/2)}^{2}}}+\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{QQ}{{{x}^{2}}}=0\] |
\[\Rightarrow \] \[q=-\frac{Q}{4}\] |
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