The figure represents two equipotential lines | vedclass

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(B) The electric field component $E_{x}$ is given by the negative gradient of the potential with respect to $x$, i.e.,$E_{x} = -\frac{dV}{dx}$.From th

Vedclass · Accepted Answer

$-100$

Vedclass · Answer

(B) The electric field component $E_{x}$ is given by the negative gradient of the potential with respect to $x$, i.e.,$E_{x} = -\frac{dV}{dx}$.From the graph, the potential difference $\Delta V = V_{2} - V_{1} = 4 \ V - 2 \ V = 2 \ V$.The distance along the $x$-axis between these two equipotential lines is $\Delta x = 4 \ cm - 2 \ cm = 2 \ cm = 0.02 \ m$.Therefore, the magnitude of the $x$-component of the electric field is $E_{x} = -\frac{\Delta V}{\Delta x} = -\frac{2 \ V}{0.02 \ m} = -100 \ V/m$.Thus, the correct option is $B$.

The figure represents two equipotential lines in the $x-y$ plane for an electric field. The $x$-component $E_{x}$ of the electric field in the space between these equipotential lines is, (in $V/m$)

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