Question:
The electric potential at points along x-direction is given by \(V = (1.0~\text{V/m}^2) x^2\). The electric field at point \(x = 5.0~\text{m}\) on x-axis is:
The electric potential at points along x-direction is given by \(V = (1.0~\text{V/m}^2) x^2\). The electric field at point \(x = 5.0~\text{m}\) on x-axis is:
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Electric field is negative gradient of potential: \( \vec{E} = - \nabla V \). For 1D along x-axis, \(E_x = - dV/dx\).
Updated On: Jun 19, 2026
- -10.0 V/m
- +10.0 V/m
- -5.0 V/m
- +5.0 V/m
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The Correct Option is A
Solution and Explanation
Step 1: Electric field from potential.
\[ E_x = - \frac{dV}{dx} \]
Step 2: Differentiate potential.
\[ V = 1.0 x^2 \Rightarrow \frac{dV}{dx} = 2 \cdot 1.0 \cdot x = 2 x \]
Step 3: Compute electric field at \(x = 5\) m.
\[ E_x = - \frac{dV}{dx} = -2 \cdot 5 = -10~\text{V/m} \]
Step 4: Conclusion.
The electric field at \(x = 5.0~\text{m}\) is -10.0 V/m.
\[ E_x = - \frac{dV}{dx} \]
Step 2: Differentiate potential.
\[ V = 1.0 x^2 \Rightarrow \frac{dV}{dx} = 2 \cdot 1.0 \cdot x = 2 x \]
Step 3: Compute electric field at \(x = 5\) m.
\[ E_x = - \frac{dV}{dx} = -2 \cdot 5 = -10~\text{V/m} \]
Step 4: Conclusion.
The electric field at \(x = 5.0~\text{m}\) is -10.0 V/m.
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