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(B) The equation of the equipotential lines is $V = ax + by + c$. From the graph,for $V = 2 \, V$,the line passes through $(6, 2)$ and $(4, 1)$. The s

Vedclass · Accepted Answer

$-100 \, V/m, +200 \, V/m$

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(B) The equation of the equipotential lines is $V = ax + by + c$. From the graph,for $V = 2 \, V$,the line passes through $(6, 2)$ and $(4, 1)$. The slope is $m = \frac{1-2}{4-6} = \frac{-1}{-2} = 0.5$. The equation is $y - 1 = 0.5(x - 4) \Rightarrow y = 0.5x - 1 \Rightarrow 0.5x - y = 1$. Multiplying by $2$,we get $x - 2y = 2$. Thus,$V = k(x - 2y)$. For $V = 2$,$2 = k(6 - 2(2)) = 2k \Rightarrow k = 1$. So,$V = x - 2y$. The electric field components are $E_x = -\frac{\partial V}{\partial x} = -1 \, V/	ext{cm} = -100 \, V/m$ and $E_y = -\frac{\partial V}{\partial y} = -(-2) = +2 \, V/	ext{cm} = +200 \, V/m$. Thus,$E_x = -100 \, V/m$ and $E_y = +200 \, V/m$.

The figure shows two equipotential lines in the $x, y$ plane for an electric field. The scales are marked in $\text{cm}$. The $x$-component $E_x$ and $y$-component $E_y$ of the electric field in the space between these equipotential lines are respectively:

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