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.
Step 1: Understanding the Concept:
Electric potential is a scalar field, and the electric field is a vector field representing its spatial rate of change.
The one-dimensional electric field is the negative derivative of the electric potential with respect to position. Step 2: Key Formula or Approach:
$E_x = -\frac{dV}{dx}$.
We differentiate the given polynomial function and evaluate it at the specified coordinate. Step 3: Detailed Explanation:
Given potential function:
\[ V(x) = 4x^2 + 8x - 3 \]
Differentiate $V$ with respect to $x$:
\[ \frac{dV}{dx} = \frac{d}{dx}(4x^2) + \frac{d}{dx}(8x) - \frac{d}{dx}(3) \]
\[ \frac{dV}{dx} = 8x + 8 - 0 \]
The electric field $E$ is the negative gradient:
\[ E = -\frac{dV}{dx} = -(8x + 8) \]
We need to find $E$ at $x = 0.5 \text{ m}$. Substitute $x = 0.5$:
\[ E = -(8(0.5) + 8) \]
\[ E = -(4 + 8) \]
\[ E = -12 \text{ V/m} \]
Step 4: Final Answer:
The electric field is $-12 \text{ V/m}$.