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Confused! kindly explain, - Electrostatics - JEE Main-7

The distance between the two charges 25\mu C  and  36\mu C  is 11 cm  At what point on the line joining the two, the intensity will be zero

  • Option 1)

    At a distance of 5 cm from 25\mu C             

  • Option 2)

    At a distance of 5 cm from 36\mu C

  • Option 3)

    At a distance of 10 cm from 25\mu C      

  • Option 4)

    At a distance of 11 cm from 36\mu C

 
Answers (1)
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V Vakul

As we learned

Electric Field Intensity -

\vec{E}=\frac{\vec{F}}{q_{0}}=\frac{kQ}{r^{2}}

- wherein

 

 Suppose electric field is zero at point N in the figure then

At N |E1| = |E2 | which gives x_{1}=\frac{x}{\sqrt{\frac{Q_2}{Q_1}}+1}=\frac{11}{\sqrt{\frac{36}{25}}+1}=5\; cm


Option 1)

At a distance of 5 cm from 25\mu C             

Option 2)

At a distance of 5 cm from 36\mu C

Option 3)

At a distance of 10 cm from 25\mu C      

Option 4)

At a distance of 11 cm from 36\mu C

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