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(C) The relationship between electric field $E$ and electric potential $V$ is given by $E = -dV/dx$. This means the magnitude of the electric field $|

Vedclass · Accepted Answer

$E_1 = E_3$ and $E_2 < E_4$

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(C) The relationship between electric field $E$ and electric potential $V$ is given by $E = -dV/dx$. This means the magnitude of the electric field $|E|$ is equal to the absolute value of the slope of the $V-x$ graph, i.e.,$|E| = |dV/dx|$.$1$. In region $1$ (segment $AB$), the potential $V$ is constant, so $dV/dx = 0$, which implies $E_1 = 0$.$2$. In region $2$ (segment $BC$), the potential increases linearly. The slope is positive, so $E_2 = -(	ext{positive value}) $3$. In region $3$ (segment $CD$), the potential $V$ is constant, so $dV/dx = 0$, which implies $E_3 = 0$.$4$. In region $4$ (segment $DE$), the potential decreases sharply. The slope is negative and steeper than in region $2$, so $E_4 = -(	ext{negative value}) > 0$. The magnitude $|E_4|$ is constant and greater than $|E_2|$.Comparing the magnitudes: $E_1 = E_3 = 0$, and $|E_4| > |E_2| > 0$. Thus, $E_1 = E_3$ and $|E_2| < |E_4|$. The correct option is $C$.

The following figure shows the electric potential $V$ as a function of position $x$ across $4$ regions on the $x$-axis. Which of the following is correct for the electric field $E$ in these regions?

Similar Questions

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