Question:
If the electric potential is given by $V = 3x^2 + 4x$ volt, then the magnitude of the electric field at the point $x = 1m$ is
If the electric potential is given by $V = 3x^2 + 4x$ volt, then the magnitude of the electric field at the point $x = 1m$ is
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Electric field is negative gradient of potential.
Updated On: Apr 24, 2026
- $6Vm^{-1}$
- $4Vm^{-1}$
- $8Vm^{-1}$
- $10Vm^{-1}$
- $12Vm^{-1}$
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The Correct Option is D
Solution and Explanation
Concept:
• Electric field: \[ E = -\frac{dV}{dx} \]
Step 1: Differentiate
\[ \frac{dV}{dx} = 6x + 4 \]
Step 2: At $x=1$
\[ E = -(6(1)+4) = -10 \] Magnitude: \[ |E| = 10 \] Final Conclusion:
Option (D)
• Electric field: \[ E = -\frac{dV}{dx} \]
Step 1: Differentiate
\[ \frac{dV}{dx} = 6x + 4 \]
Step 2: At $x=1$
\[ E = -(6(1)+4) = -10 \] Magnitude: \[ |E| = 10 \] Final Conclusion:
Option (D)
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