Two charges q₁ and q₂ are separated by a distance of 30cm. A third charge q₃ initially at point C is shown in figure, is moved along the circular path of radius 40 cm from C to D. If the difference in potential energy due to movement of q₃ from C to D is q₃ k/4 ₀ , the value of K is ( ₀ = permittivity of free space)

Step-by-step solution

The initial potential energy of the system when charge q₃ is at point C is given by: U_C = 1 4 ₀ ( q₁ q₃ AC + q₂ q₃ BC ) From the given figure, AC = 40 cm = 0.4 m and AB = 30 cm = 0.3 m . In the right-angled triangle ABC, the distance BC is: BC = AB^2 + AC^2 = (0.3)^2 + (0.4)^2 = 0.5 m So, U_C = q₃ 4 ₀ ( q₁ 0.4 + q₂ 0.5 ) When q₃ is moved to point D along the circular path centered at A, the distance AD is equal to the radius of the path: AD = AC = 40 cm = 0.4 m The distance BD is: BD = AD - AB = 0.4 m - 0.3 m = 0.