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Two equal charges of magnitude q each are placed at a distance d apart.Their electrostatic energy is E.A third charge-Q/2 is brought midway between these two charges.The electrostatic energy of this system is now

AKASH Das , 6 Years ago
Grade 11
anser 1 Answers
harshin nechooli

Last Activity: 6 Years ago

Here the energy refers to the electrostatic potential energy does initially there have a potential energy of k q bhai d then when charge is brought final potential energy after system will change to 5 e
finally   PE = kq÷(d÷2) +kq÷(d÷2) +E. and E= kq÷d

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