Question:
The electric potential '$V$' is given as a function of distance '$x$' (metre) by $V = (4x^2 + 8x - 3)V$. The value of electric field at $x = 0.5 \text{ m}$, in $\text{V/m}$ is
The electric potential '$V$' is given as a function of distance '$x$' (metre) by $V = (4x^2 + 8x - 3)V$. The value of electric field at $x = 0.5 \text{ m}$, in $\text{V/m}$ is
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Electric field points in the direction of decreasing potential.
Updated On: May 12, 2026
- -16
- -12
- 0
- +12
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The Correct Option is B
Solution and Explanation
Step 1: Concept
Electric field is the negative gradient of potential: $E = -dV/dx$.
Step 2: Meaning
$dV/dx = \frac{d}{dx}(4x^2 + 8x - 3) = 8x + 8$.
Step 3: Analysis
At $x = 0.5 \text{ m}$: $dV/dx = 8(0.5) + 8 = 4 + 8 = 12$. $E = -(12) = -12 \text{ V/m}$.
Step 4: Conclusion
The electric field value is $-12 \text{ V/m}$. Final Answer: (B)
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