The electric potential V at any point (x, y, z) in space is given by \( V=6x\,\mathrm{V} \), where x is in metres. The electric field at the point (1 m, 0, 2 m) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( -6\,\hat{\imath}\,\mathrm{V/m} \)
The electric field is the negative rate of change of potential with distance, so it points in the direction in which the potential decreases. Here \( V \) changes only with \( x \), so the field has only an x-component.
\( \begin{aligned} E_x &= -\frac{dV}{dx} \\ &= -\frac{d(6x)}{dx}\,\mathrm{V/m} \\ &= -6\,\mathrm{V/m} \end{aligned} \)
Since \( V \) does not depend on \( y \) or \( z \), the components \( E_y \) and \( E_z \) are zero. Therefore,
\( \vec{E}=-6\,\hat{\imath}\,\mathrm{V/m} \)
The field is the same at every point, including (1 m, 0, 2 m), and points along −x, where \( V \) falls.
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.”