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(B) The electric field $E$ due to an infinite nonconducting sheet of charge is given by $E = \frac{\sigma}{2\varepsilon_0}$.The potential difference $

Vedclass · Accepted Answer

$0.88 \ mm$

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(B) The electric field $E$ due to an infinite nonconducting sheet of charge is given by $E = \frac{\sigma}{2\varepsilon_0}$.The potential difference $V$ between two points separated by a distance $d$ in a uniform electric field is $V = E \cdot d$.Substituting the expression for $E$,we get $V = \frac{\sigma}{2\varepsilon_0} \cdot d$.Rearranging for $d$,we have $d = \frac{2\varepsilon_0 V}{\sigma}$.Given $\sigma = 10^{-7} \ C/m^2$,$V = 5 \ V$,and $\varepsilon_0 = 8.854 	imes 10^{-12} \ C^2/(N \cdot m^2)$.$d = \frac{2 	imes 8.854 	imes 10^{-12} 	imes 5}{10^{-7}} = 88.54 	imes 10^{-5} \ m = 0.8854 	imes 10^{-3} \ m = 0.88 \ mm$.

An infinite nonconducting sheet of charge has a surface charge density of $10^{-7} \ C/m^2$. The separation between two equipotential surfaces near the sheet whose potential differ by $5 \ V$ is:

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