Electric potential increases along the x-axis. A proton and an electron are placed at rest at the origin.
In which direction do the charges move when released?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → The proton moves towards the negative x-axis, while the electron moves towards the positive x-axis.
The electric field points in the direction in which the potential decreases. Since the potential rises along +x, the field \( \vec{E} \) at the origin is along the negative x-axis. A charge released from rest moves along the force \( \vec{F}=q\vec{E} \), so a positive charge moves along \( \vec{E} \) and a negative charge moves opposite to it.
Option (1) The proton moves towards the negative x-axis, while the electron moves towards the positive x-axis. — Correct. The proton (\( +e \)) feels the force \( e\vec{E} \) along −x, towards lower potential. The electron (\( -e \)) feels \( -e\vec{E} \) along +x, towards higher potential.
Option (2) The proton moves towards the positive x-axis, while the electron moves towards the negative x-axis. — Incorrect. This needs a field along +x, that is, a potential that falls along +x. Here the potential rises along +x.
Option (3) The proton and the electron both move towards the negative x-axis. — Incorrect. The electron carries negative charge, so the same field pushes it opposite to the proton, along +x.
Option (4) The proton and the electron both move towards the positive x-axis. — Incorrect. The proton is a positive charge, so it is pushed towards lower potential, which lies along −x.
NCERT (2026-27):
Relation Between Field and Potential – “We thus arrive at two important conclusions concerning the relation between electric field and potential: (i) Electric field is in the direction in which the potential decreases steepest.”