Two charges q₁ and q₂ are separated by a distance of 30 cm . A third charge q₃ initially at ' C ' as shown in the 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 given by (q₃ K)/(4 π ε₀), the value of K is :

Solution

Physics sir · 3 steps

q1 sits at the centre of the circle, so only q2's distance to q3 changes. Track that one distance.

1. q1 drops out. The path is a circle of radius 40 cm centred at A, where q₁ sits. q₃ stays 0.4 m from q₁ the whole way, so the energy shared by q₁ and q₃ does not change.
2. Distances from q2. At C, BC = √(0.3² + 0.4²) = 0.5 m (a 3-4-5 triangle). At D, BD = 0.4 - 0.3 = 0.1 m.

   Δ U = (q₂ q₃)/4πε₀(1/0.1 - 1/0.5) = q₃/4πε₀(10 - 2)q₂

3. So. Δ U = (q₃ × 8q₂)/4πε₀, so K = 8q₂. That is option (d).

The idea to keep. Potential energy depends only on distances. Any charge whose distance to the moving charge stays fixed adds nothing to the change, so find it first and ignore it.

Answer from the NTA final answer key. Solution by PrepWiser.