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MHT CET202312 May 2023Evening ShiftPhysicsElectrostaticsActual

Two point charges ' q 1 ' and 'q2' are separated by a distance ' d '. What is the increase in potential energy of the system when ' q 2 ' is moved towards ' q 1 ' by a distance ' x '? ( x < d ) ( 1 4 ₀ = K , constant )

Options

  1. A- Kq ₁ q ₂ x d ( d - x )
  2. B- Kq ₁ q ₂ ~d ( ~d - x )
  3. CKq ₁ q _ 2 x ( ~d ^2- x ^2 )
  4. DKq ₁ q ₂ x ( d ^2- x ^2 )

Correct answer

A. - Kq ₁ q ₂ x d ( d - x )

Step-by-step solution

The potential energy between two charges is given as U= k q₁ q₂ r Initial potential energy is U _ f = kq ₁ q ₂ r When charge q₂ moves towards the q₁ the separation between the charges becomes d - x The final potential energy is U _ f = kq ₁ q ₂ ( ~d - x ) The increase in potential energy is array ll & U=U_f-U_f & U= k q₁ q₂ d - k q₁ q₂ (d-x) & U=k q₁ q₂ ( 1 d - 1 d-x ) & U= -k q₁ q₂ x d(d-x) array

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