A) 1 cm
B) 2 cm
C) 3cm
D) 5cm
Correct Answer: A
Solution :
Let \[P\] be a point at a distance x from charge\[-q\] at which electric field \[E=0.\] For no field at P, \[{{E}_{1}}={{E}_{2.}}\] or\[\frac{1}{2\pi {{\varepsilon }_{0}}}\frac{q}{{{x}^{2}}\times {{10}^{-4}}}=\frac{1}{2\pi {{\varepsilon }_{0}}}\frac{4q}{{{(1+x)}^{2}}\times {{10}^{-4}}}\] \[\Rightarrow \,\,\frac{1}{{{x}^{2}}}=\frac{4}{{{(1+x)}^{2}}}\] \[\Rightarrow \,\,\frac{1}{x}=\frac{2}{1+x}\] \[\therefore \] \[2x=1+x\] or \[x=1\,cm\]You need to login to perform this action.
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