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In the first configuration (1) as shown in the figure, four identical charges ( (q₀ ) ) are kept at the corners (A, B, C ) and (D ) of square of side length 'a'. In the second configuration (2), the same charges are shifted to mid points ( G , E , H ) and F, of the square, If ( K = 1 4 ₀ ), the difference between the potential energies of configuration (2) and (1) is given by :

Options

  1. AK q₀^2 a (4-2 2 )
  2. BK q₀^2 a (3- 2 )
  3. CKq ₀^2 a (4 2 -2)
  4. DKq ₀^2 a (3 2 -2)

Correct answer

D. Kq ₀^2 a (3 2 -2)

Step-by-step solution

aligned & u_ = (2 K q₀ a + K q₀ 2 a ) q₀ 2 & u₀= (2 K q₀ 2 a + K q₀ a ) q₀ 2 aligned aligned & So, u=u₂-u₁=2 q₀ k q₀ a [2 2 +1-2- 1 2 ] & u= 2 q₀^2 4 ₀ a [ 4- 2 -1 2 ]= 2 q₀^2 4 ₀ a (3- 2 ) 2 & u= 2 k q₀^2 a [ 3- 2 2 ]= k q₀^2 a (3 2 -2) aligned

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