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A point charge is surrounded symmetrically by six identical charges at distance r as shown in the figure.  How much work is done by the forces of electrostatic repulsion when the point charge q at the centre is removed at infinity 

  • Option 1)

    Zero

  • Option 2)

    \frac{6q^2}{4\pi\epsilon_0 r}

  • Option 3)

    \frac{q^2}{4\pi\epsilon_0 r}

  • Option 4)

    \frac{12q^2}{4\pi\epsilon_0 r}

 

Answers (1)

As we have learnt,

 

Electron Volt -

It is the smallest practical unit of energy if used in atomic and nuclear Physics.

 

 

 

- wherein

1\, e v= 1.6\times 10^{-19}J

1.6\times 10^{-12}erg.

 

Total potential at the centre V = \frac{6q}{4\pi\epsilon_0 r}

Required work done = q\cdot V = \frac{6q^2}{4\pi\epsilon_0 r} 

 


Option 1)

Zero

Option 2)

\frac{6q^2}{4\pi\epsilon_0 r}

Option 3)

\frac{q^2}{4\pi\epsilon_0 r}

Option 4)

\frac{12q^2}{4\pi\epsilon_0 r}

Posted by

Vakul

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