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(D) The electric field $E$ produced by an infinite non-conducting sheet is given by $E = \frac{\sigma}{2 \varepsilon_0}$.Given $\sigma = 7 	imes 10^{

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$0.5$

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(D) The electric field $E$ produced by an infinite non-conducting sheet is given by $E = \frac{\sigma}{2 \varepsilon_0}$.Given $\sigma = 7 	imes 10^{-7} 	ext{ C m}^{-2}$ and $\frac{1}{4 \pi \varepsilon_0} = 9 	imes 10^9 	ext{ SI units}$.We know that $\varepsilon_0 = \frac{1}{4 \pi 	imes 9 	imes 10^9} \approx 8.85 	imes 10^{-12} 	ext{ F m}^{-1}$.The relation between electric field and potential difference $\Delta V$ for a distance $\Delta r$ is $|E| = \frac{\Delta V}{\Delta r}$.Substituting the values:$\frac{\sigma}{2 \varepsilon_0} = \frac{\Delta V}{\Delta r}$$\Delta r = \frac{\Delta V 	imes 2 \varepsilon_0}{\sigma} = \frac{19.8 	imes 2 	imes 8.85 	imes 10^{-12}}{7 	imes 10^{-7}}$$\Delta r = \frac{19.8 	imes 17.7 	imes 10^{-12}}{7 	imes 10^{-7}} \approx 5 	imes 10^{-4} 	ext{ m} = 0.5 	ext{ mm}$.

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