KCET · Physics · Electrostatics
A uniform electric field vector \(\mathbf{E}\) 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
- A \(q\left[V_A-E x\right]\)
- B \(q\left[V_A+E x\right]\)
- C \(q\left[E x-V_A\right]\)
- D \(q E x\)
Answer & Solution
Correct Answer
(A) \(q\left[V_A-E x\right]\)
Step-by-step Solution
Detailed explanation
Potential energy at \(B=\) Potential difference \(\times\)
Charge + Energy spent by moving charge from \(A\) to \(B\)
\(=V_A \times q-q \cdot E \cdot x=q\left(V_A-E x\right)\)
Charge + Energy spent by moving charge from \(A\) to \(B\)
\(=V_A \times q-q \cdot E \cdot x=q\left(V_A-E x\right)\)
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