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  • Answer please! A charge 10 e.s.u. is placed at a distance of 2 cm from a charge 40 e.s.u. and 4 cm from another charge of 20 e.s.u. The potential energy of the charge 10 e.s.u. is (in ergs)

A charge 10 e.s.u. is placed at a distance of 2 cm from a charge 40 e.s.u. and 4 cm from another charge of 20 e.s.u. The potential energy of the charge 10 e.s.u. is (in ergs)

  • Option 1)

    87.5

  • Option 2)

    112.5

  • Option 3)

    150

  • Option 4)

    250

 

Answers (1)

As we have learnt,

 

Potential Energy Of System Of two Charge -

U=\frac{kQ_{1}Q_{2}}{r}  \left ( S.I \right )

U= \frac{Q_{1}Q_{2}}{r}  \left ( C.G.S \right ) 

 

- wherein

K=\frac{1}{4\pi \epsilon _{0}}

Potential energy of 10 e.s.u. charge is

U = \frac{10\times 40}{2} + \frac{10\times 20}{4} = 250erg 

 


Option 1)

87.5

Option 2)

112.5

Option 3)

150

Option 4)

250

Posted by

Vakul

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