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  • Can someone help me with this, - Electrostatics - JEE Main-7

Two charges +3.2\times 10^{-19} and -3.2\times 10^{-19}  kept 2.4 Å apart forms a dipole. If it is kept in uniform electric field of intensity 4\times 10^5 volt/m  then what will be its electrical energy in equilibrium

  • Option 1)

    +3\times 10^{-23}J

  • Option 2)

    -3\times 10^{-23}J

  • Option 3)

    -6\times 10^{-23}J

  • Option 4)

    -2\times 10^{-23}J

 

Answers (1)

best_answer

As we have learned

Stable Equilibriun -

\Theta = 90^{\circ}

\tau =0

w = 0

U_{min} = -PE

- wherein

 

 

Potential energy of electric dipole

U=-pE \cos \theta = -(q\times 2l)E\cos \theta 

U=-(3.2*10^{-19}*2.4*10^{-10})4*10^5\cos \theta

U=-(3*10^{-23})  (approx.)

 


Option 1)

+3\times 10^{-23}J

Option 2)

-3\times 10^{-23}J

Option 3)

-6\times 10^{-23}J

Option 4)

-2\times 10^{-23}J

Posted by

prateek

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