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Three charges Q, +q and +q are placed at vertices of a right - angle isosceles triangle as shown blow. The net electrostatic energy of the configuration is zero, if the value of Q is:

  • Option 1)

    \frac{-q}{1+\sqrt{2}}

  • Option 2)

    \frac{-\sqrt{2}q}{\sqrt{2}+1}

  • Option 3)

    +q

  • Option 4)

    -2q

Answers (1)

best_answer

 

Potential Energy Of a System Of n Charge -

U= K\left ( \frac{Q_{1}Q_{2}}{r_{12}} +\frac{Q_{2}Q_{3}}{r_{23}}+\frac{Q_{1}Q_{3}}{r_{13}}\right )

- wherein

For system of 3 charges.

Energy of photon 

E=\frac{hc}{\lambda }=\frac{12500}{980}=12.75 ev

Since D_{m}=(M-1)A and as wave length increases M decreases, so Dm decreases

 

\becauseelectron will be exite to n=4 since R\alpha n^{2}

\because Radius of atom will be 16 ao.

U=K\; \; \; K\left [ \frac{q^{2}}{d}+\frac{Qq}{d}+\frac{Qq}{d\sqrt{2}} \right ]=0

\Rightarrow Q=\frac{-q\sqrt{2}}{\sqrt{2}+1}

 

 


Option 1)

\frac{-q}{1+\sqrt{2}}

Option 2)

\frac{-\sqrt{2}q}{\sqrt{2}+1}

Option 3)

+q

Option 4)

-2q

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