A very large plastic sheet carries a uniform charge density of $\mathbf{-}\mathbf{6}\mathbf{.}\mathbf{00}\mathit{n}\mathit{C}\mathbf{/}{\mathit{m}}^{\mathbf{2}}$on one face. (a) As you move away from the sheet along a line perpendicular to it, does the potential increase or decrease? How do you know, without doing any calculations? Does your answer depend on where you choose the reference point for potential? (b) Find the spacing between equipotential surfaces that differ from each other by 1.0

V. What type of surfaces are these?

The charge of the sheet is negative as given, which indicates that the direction of the electric field is towards the charged sheet, and moving in the direction of the electric field means moving in the direction of decreasing potential.

Therefore, as to move in the opposite direction of the electric field the potential increases.

For the second question, it does not depend on a reference point for the potential. In anyway, to work to move away from the negative sheet and as to do work, the potential will increase as to move away from the sheet.

The potential between the two surfaces$\mathbf{\nabla }\mathit{V}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{00}\mathit{V}$. also the electric field is related to the potential difference by equation in the form

${\mathit{V}}_{\mathbf{a}}\mathbf{-}{\mathit{V}}_{\mathbf{b}}\mathbf{=}\mathbf{∆}\mathit{V}\mathbf{=}\mathbf{-}\mathit{\xi }\mathbf{}\mathit{E}\mathbf{.}\mathbf{}\mathit{d}\mathit{I}$

As the distance field E for an infinite charged sheet could be obtained from example 22.7 in the next form

localid="1664267858164" $\mathit{E}\mathbf{=}\frac{\mathbf{\sigma }}{\mathbf{2}\mathbf{\in }{}_{\mathbf{0}}}$

Where $\in {}_0$ is the electric constant and equals to $8.854\times {10}^{-12}{C}^2/N.m^2$and $\sigma$ is the charge density of the sheet.

Expressing E in relation to complete the calculation

$$\begin{gathered} I=\frac{-∆V}{E} \\ I=\frac{-2{\epsilon }_0∆V}{}\sigma \end{gathered}$$

Now plug the values into the equation (1) to get l,

$$\begin{gathered} I=\frac{-2{\epsilon }_0∆V}{\sigma } \\ =\frac{-2\left(8.854\times {10}^{-12}{C}^2/N.m\right)\left(1.00V\right)}{\left(-6.00*{10}^{-9}C/m\right)} \\ =0.003m \\ =0.30cm \end{gathered}$$

The electric field is perpendicular to the sheet, so the equipotential surfaces are parallel planes to the sheet. The electric field is perpendicular to equipotential.