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  • Can someone help me with this, - Electrostatics - JEE Main-5

What is the magnitude of a point charge due to which the electric field 30 cm away has the magnitude 2 newton/coulomb [1/4\pi \varepsilon _{0}=9\times 10^{9}Nm^{2}]

  • Option 1)

    2\times 10^{-11} coulomb

     

     

     

  • Option 2)

    3\times 10^{-11} coulomb

  • Option 3)

    5\times 10^{-11} coulomb

  • Option 4)

    9\times 10^{-11} coulomb

 

Answers (1)

best_answer

As we learned

 

Electric field and potential Due to various charge Distribution i) Point Charge -

Electric field and Potential at Point P due to a Point charge Q.

E= \frac{kQ}{r^{2}}    ,     V= \frac{kQ}{r}

- wherein

 

 By using E=\frac{1}{4\pi \varepsilon _{0}}\cdot \frac{Q}{r^{2}};            2=9\times 10^{9}\times \frac{Q}{30\times 10^{-2}}\Rightarrow Q=2\times 10^{-11}C


Option 1)

2\times 10^{-11} coulomb

 

 

 

Option 2)

3\times 10^{-11} coulomb

Option 3)

5\times 10^{-11} coulomb

Option 4)

9\times 10^{-11} coulomb

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