The given figure shows the variation of electric potential V as a fun…

The correct answer is Option (2)

The electric field is the negative gradient of the potential, \( E = -dV/dx \). Read the slope of the V-x graph in each region.

    \( 0 \text{ to } 2: \text{ V rises linearly, slope positive} \Rightarrow E = -\text{(constant)} \lt 0 \)

    \( 2 \text{ to } 4: \text{ V is constant, slope zero} \Rightarrow E = 0 \)

    \( 4 \text{ to } 6: \text{ V falls linearly, slope negative} \Rightarrow E = +\text{(constant)} \gt 0 \)

So the E-x graph has a negative constant value from 0 to 2, zero from 2 to 4, and a positive constant value from 4 to 6. This is the graph of Option (2).

Why the other options are incorrect

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  Option (1) — Incorrect. The field is shown positive and constant right up to 4; the flat part of the potential (2 to 4) must give zero field, and the first region must give a negative field.

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  Option (3) — Incorrect. The signs are reversed, as if \( E = +dV/dx \); the field must be negative where V rises and positive where V falls.

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  Option (4) — Incorrect. A linearly varying field would need a potential that varies as the square of the distance, but the potential here is linear, so the field is constant in each region.

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. (ii) Its magnitude is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point.”