A narrow beam contains two particles, A and B

PHYSICS · Multi-part · CBSE · Physics · Electrostatic Potential and Capacitance · Electrostatic Potential and Capacitance

Problem

A narrow beam contains two particles, A and B, that travel simultaneously along the same straight line in a region where the electric potential V(x)V(x) varies with position as described below: • V(x)V(x) increases linearly from 0 V0\,\text{V} at x=0 mx = 0\,\text{m} to 40 V40\,\text{V} at x=2 mx = 2\,\text{m}. • V(x)V(x) then decreases quadratically, following V(x)=40 V−5 (x−2 m)2V(x) = 40\,\text{V} - 5\,(x - 2\,\text{m})^{2}, until x=4 mx = 4\,\text{m}. Particle A carries charge +2q+2q and mass 2m2m. Particle B carries charge −q-q and mass mm. At x=0 mx = 0\,\text{m}, both particles are released from rest. Neglect all forces other than the electrostatic force. (a) Derive an expression for the electric potential energy of each particle as a function of position in the interval 0 m≤x≤2 m0\,\text{m} \le x \le 2\,\text{m}. (b) Determine which particle has the greater speed at x=2 mx = 2\,\text{m}. Provide a quantitative comparison supported by calculations. (c) Between x=2 mx = 2\,\text{m} and x=4 mx = 4\,\text{m}, determine whether either particle comes to rest (turning point). If so, calculate the position(s) where this occurs and identify which particle(s) are affected. (d) Using energy arguments, discuss whether the two particles could ever have the same speed at some position x>0 mx > 0\,\text{m}. If so, state at least one possible position and justify; if not, explain why this is impossible.

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