Question

the electric potential at a certain point is given by v(x)=139x² + 41.0x, where v is in volts and x is in meters. what is the x - component of the electric field at x = - 2.09 m?

Explanation:

Step1: Recall the relationship between electric - field and potential

The relationship between the electric - field component $E_x$ and the electric potential $V(x)$ is $E_x=-\frac{dV}{dx}$. Given $V(x) = 139x^{2}+41.0x$.

Step2: Differentiate the potential function

Using the power - rule for differentiation $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we have $\frac{dV}{dx}=\frac{d}{dx}(139x^{2}+41.0x)$.
$\frac{dV}{dx}=2\times139x + 41.0=278x + 41.0$.
Then $E_x=-(278x + 41.0)$.

Step3: Solve for $x$ when $E_x=-2.09$ V/m

Set $-2.09=-(278x + 41.0)$.
First, cancel out the negative signs on both sides: $2.09 = 278x+41.0$.
Then, rearrange the equation to solve for $x$: $278x=2.09 - 41.0$.
$278x=-38.91$.
$x=\frac{-38.91}{278}\approx - 0.14$ m.

Answer:

$x\approx - 0.14$ m