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KCET2023PhysicsElectrostatics

A uniform electric field vector E exists along horizontal direction as shown. The electric potential at A is V_A . A small point charge q is slowly taken from A to B along the curved path as shown. The potential energy of the charge when it is at point B is

Options

  1. Aq [V_A-E x ]
  2. Bq [V_A+E x ]
  3. Cq [E x-V_A ]
  4. Dq E x

Correct answer

A. q [V_A-E x ]

Step-by-step solution

Potential energy at B= Potential difference Charge + Energy spent by moving charge from A to B=V_A q-q E x=q (V_A-E x )

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