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(C) The relationship between electric field $E$ and potential difference $dV$ is given by $dV = -\vec{E} \cdot d\vec{r} = -E \, dr \cos 	heta$.Here,t

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$200$

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(C) The relationship between electric field $E$ and potential difference $dV$ is given by $dV = -\vec{E} \cdot d\vec{r} = -E \, dr \cos 	heta$.Here,the potential increases in the direction of the electric field. The distance between the equipotential lines of $20 \, V$ and $30 \, V$ along the $x$-axis is $dr = (20 - 10) \, cm = 10 	imes 10^{-2} \, m$.The angle between the electric field vector $\vec{E}$ and the displacement vector $d\vec{r}$ along the $x$-axis is $	heta = 180^\circ - 30^\circ = 150^\circ$ (or using the perpendicular distance method,$E = \frac{dV}{dr_{\perp}}$ where $dr_{\perp} = dr \sin 30^\circ$).Using $E = \frac{\Delta V}{\Delta r \sin 	heta} = \frac{30 - 20}{0.1 	imes \sin 30^\circ} = \frac{10}{0.1 	imes 0.5} = \frac{10}{0.05} = 200 \, V/m$.

For the given figure,the value of the electric field is ......... $V/m$.

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