A) 22.5cm
B) 18cm
C) 36 cm
D) 30 cm
Correct Answer: D
Solution :
Let a test charge \[{{q}_{0}}\] is placed at a point D at a distance \[x\]from point A. From Coulombs law, the force on the test charge \[{{q}_{0}}\] due to a charge q, kept a distance r apart is \[F=-\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{q\,{{q}_{0}}}{{{r}^{2}}}\] \[\therefore \] Electric field intensity at the point D \[E=\frac{F}{4\,\pi {{\varepsilon }_{0}}}=\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}\frac{q\,{{q}_{0}}}{{{r}^{2}}}\] Let intensity at D distance x from point A be zero, then \[\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}.\frac{10\times {{10}^{-6}}}{{{x}^{2}}}=\frac{1}{4\,\pi \,{{\varepsilon }_{0}}}\frac{40\times {{10}^{-6}}}{{{(90-x)}^{2}}}\] \[\Rightarrow \] \[\frac{1}{{{x}^{2}}}=\frac{4}{{{(90-x)}^{2}}}\] \[\Rightarrow \] \[\frac{1}{x}=\frac{2}{90-x}\] \[\Rightarrow \] \[90-x=2x\] \[\Rightarrow \] \[3x=90\] \[\Rightarrow \] \[x=30\,cm\]. Note: The force F acts along the line joining the two charges.You need to login to perform this action.
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