question_answer

  Two small charged spheres A and B have charges \[10\,\mu \,C\] and \[40\,\mu \,C\], respectively and are held at a separation of 90 cm from each other At what distance from A, electric intensity would be zero?

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.