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A charge q is placed at the centre of the line joining two equal charges Q. The system of the three charges will be in equilibrium. If q is equal to 

  • Option 1)

    -\frac{Q}{2}

  • Option 2)

    -\frac{Q}{4}

  • Option 3)

    +\frac{Q}{4}

  • Option 4)

    +\frac{Q}{2}

 

Answers (1)

best_answer

As we have learnt,

 

Neutral Point /Zero Electric Field -

Due to a system of two like Point Ooint charge.

\dpi{100} x_{1}= \frac{x}{\sqrt{\frac{Q_{2}}{Q_{1}}}+1}       ,           x_{2}= \frac{x}{\sqrt{\frac{Q_{1}}{Q_{2}}}+1}

- wherein

Where, x = distance between Q1 and Q2

 

 By using Tricky formula

q = Q\left(\frac{\frac{x}{2}}{x} \right )^2 \Rightarrow q =\frac{Q}{4}

Snce q should be negetive, so q =-\frac{Q}{4}

 


Option 1)

-\frac{Q}{2}

Option 2)

-\frac{Q}{4}

Option 3)

+\frac{Q}{4}

Option 4)

+\frac{Q}{2}

Posted by

Avinash

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