Charges q₁ and q₂ are separated by 30 cm. A third charge q₃ is moved from C to D along the circular arc of radius 40 cm shown in the original figure. If the change in potential energy is ΔU = q₃K/(4πε₀), then K is:

Question

G1 Physics · Accepted Answer

From the geometry, charge q₃ remains 0.4 m from q₁ at both C and D, so the potential-energy contribution due to q₁ does not change. Distance from q₂: At C, r C r_C r C ​ = 0.5 m. At D, r D r_D r D ​ = 0.1 m. Potential difference due to q₂ is V D − V C V_D-V_C V D ​ − V C ​ = (1/4πε₀)q₂(1/0.1 - 1/0.5) = (1/4πε₀)q₂(10-2) = (1/4πε₀)(8q₂). Thus, ΔU = q₃( V D − V C ) V_D-V_C) V D ​ − V C ​ ) = q₃(8q₂)/(4πε₀). Hence K = 8q₂. Correct option: A.