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  • How to solve this problem- - Electrostatics - JEE Main-2

Equal charges of  \frac{10}{3}\times 10^{-9}  coulomb are lying on the corners of a square of side 8 cm. The electric potential at the point of intersection of the diagonals will be

  • Option 1)

    900V

     

     

     

  • Option 2)

    900\sqrt{2}V

  • Option 3)

    150\sqrt{2}V

  • Option 4)

    1500\sqrt{2}V

 

Answers (1)

best_answer

As we learned

 

Potential of a System of Charge -

V=\frac{kq_{1}}{r_{1}}+k\frac{q_{2}}{r_{2}}+\frac{k\left ( -q_{3} \right )}{r_{3}}+\cdots

- wherein

 

 

 

Potential at the centre O 

V=4\times \frac{1}{4\pi \varepsilon _{0}}\cdot \frac{Q}{a/\sqrt{2}}given Q=\frac{10}{3}\times 10^{-9}C\Rightarrow a=8cm=8\times 10^{-2}m

V=5\times9\times 10^{9}\times \frac{\frac{10}{3\times 10^{-9}}}{\frac{8\times 10^{-2}}{\sqrt{2}}}= 1500\sqrt{2}volt

 


Option 1)

900V

 

 

 

Option 2)

900\sqrt{2}V

Option 3)

150\sqrt{2}V

Option 4)

1500\sqrt{2}V

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