2. Solved Papers
  3. RAJASTHAN PMT

RAJASTHAN PMT Rajasthan - PMT Solved Paper-2003

  • question_answer
    Two charges of magnitude\[-q\]and\[+4q\] are placed 1 cm apart. The distance between the point on the line joining and point at which the electric field is zero (with respect to\[-q\]), is

    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.
You will be redirected in 3 sec spinner