The variation of potential with distance $x$ is as shown in figure. The field at $x = 6 m$ is

- 1.5 V / m

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+ 3 V / m

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- 3 V / m

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+ 1.5 V / m

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Solution

The correct option is B $+ 3 V / m$
$E =$ -(potential gradient)
From $x = 4$ to $x = 8$
$E = - (\frac{Δ V}{Δ r}) = - (\frac{0 - 12}{4}) = \frac{12}{4} = + 3 V / m$

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