An electric field in x-y plane is given by the equation \( \vec{E} = E_{0}\,\hat{\imath}\;\mathrm{N/C} \). A positively charged particle is given a velocity of \( 5\,\hat{\jmath}\;\mathrm{m/s} \) at point x = 3, y = 0 in x-y plane. The particle will experience force :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → along x-axis
The force on a charge in an electric field is \( \vec{F} = q\vec{E} \); it does not depend on the velocity of the charge. A velocity-dependent force \( q\,\vec{v}\times\vec{B} \) would need a magnetic field, and none is given.
\( \displaystyle \vec{F} = q\vec{E} = qE_{0}\,\hat{\imath} \)
For a positive \( q \) the force is along \( +\hat{\imath} \), that is, along the x-axis, whatever the velocity or position.
Option (1) along x-axis — Correct. \( \vec{F} = qE_0\hat{\imath} \) is parallel to the field, along the x-axis.
Option (2) along y-axis — Incorrect. The y-direction is the direction of the velocity, and an electric force does not depend on the velocity.
Option (3) along z-axis — Incorrect. A force along \( z \) would come from \( \vec{v}\times\vec{B} \) with a magnetic field; there is no magnetic field here.
Option (4) along negative z-axis — Incorrect. Same reason: without a magnetic field there is no force perpendicular to both \( \vec{v} \) and \( \vec{E} \).
NCERT (2026-27):
Magnetic Field, Lorentz Force – “Let us suppose that there is a point charge q (moving with a velocity v and, located at r at a given time t) in presence of both the electric field E (r) and the magnetic field B (r). The force on an electric charge q due to both of them can be written as
\( \displaystyle \mathbf{F} = q\left[\mathbf{E}(\mathbf{r}) + \mathbf{v}\times\mathbf{B}(\mathbf{r})\right] \equiv \mathbf{F}_{\text{elec}} + \mathbf{F}_{\text{magnetic}} \)
This force was given first by H.A. Lorentz based on the extensive experiments of Ampere and others. It is called the Lorentz force. You have already studied in detail the force due to the electric field.”