The equation of an equipotential line in an e | vedclass

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Question

(D) The electric field vector $\vec{E}$ is always perpendicular to the equipotential line.The equation of the equipotential line is $y = 2x$.The slope

Vedclass · Accepted Answer

$-8 \hat{i} + 4 \hat{j}$

Vedclass · Answer

(D) The electric field vector $\vec{E}$ is always perpendicular to the equipotential line.The equation of the equipotential line is $y = 2x$.The slope of this line is $m_1 = \frac{dy}{dx} = 2$.Since the electric field vector $\vec{E} = E_x \hat{i} + E_y \hat{j}$ is perpendicular to the line,its slope $m_2$ must satisfy $m_1 \cdot m_2 = -1$.Therefore,$m_2 = -\frac{1}{2}$.The slope of the electric field vector is given by $\frac{E_y}{E_x} = -\frac{1}{2}$.Checking the options:For option $D$,$\vec{E} = -8 \hat{i} + 4 \hat{j}$,the slope is $\frac{4}{-8} = -\frac{1}{2}$.Thus,the electric field vector can be $-8 \hat{i} + 4 \hat{j}$.

The equation of an equipotential line in an electric field is $y = 2x$. The electric field strength vector at $(1, 2)$ may be:

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