Question:
The electric potential 'V' as a function of distance x (in metre) is \(V=4x^{2}+6x-8\) volt. Electric field at \(x=2\) m is
The electric potential 'V' as a function of distance x (in metre) is \(V=4x^{2}+6x-8\) volt. Electric field at \(x=2\) m is
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The field is minus the derivative of potential with respect to x.
Updated On: Nov 12, 2026
- \(-11\ \text{V/m}\)
- \(-22\ \text{V/m}\)
- \(-16\ \text{Vm}\)
- \(-8\ \text{V/m}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
\(E=-\dfrac{dV}{dx}\)
Step 2: Differentiate
\(V=4x^{2}+6x-8\Rightarrow\dfrac{dV}{dx}=8x+6\)
Step 3: Evaluate at x = 2 m
\(E=-(8\times2+6)=-22\ \text{V/m}\)
Final Answer:
The electric field at x = 2 m is minus 22 V/m. \[ \boxed{\text{(B) }-22\ \text{V/m}} \]
\(E=-\dfrac{dV}{dx}\)
Step 2: Differentiate
\(V=4x^{2}+6x-8\Rightarrow\dfrac{dV}{dx}=8x+6\)
Step 3: Evaluate at x = 2 m
\(E=-(8\times2+6)=-22\ \text{V/m}\)
Final Answer:
The electric field at x = 2 m is minus 22 V/m. \[ \boxed{\text{(B) }-22\ \text{V/m}} \]
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