91Ó°ÊÓ

The electric field of infinite parallel plates is given by:

$\mathit{E}\mathbf{=}\frac{\mathbf{σ}}{\mathbf{2}{\mathbf{Î\mu }}_{\mathbf{o}}}$ (1)

Electric potential between parallel plates at any point is given by:

$\mathit{V}\mathbf{}\mathbf{=}\mathbf{}\mathit{E}\mathit{d}$ (2)

Where $\mathit{σ}\mathbf{=}$ surface charge density,$\mathit{E}\mathbf{}\mathbf{=}\mathbf{}$ electric field,$\mathit{d}\mathbf{}\mathbf{=}\mathbf{}$ the separation between the plates, and $\mathit{V}\mathbf{}\mathbf{=}\mathbf{}$potential difference.

(a)

The electric field of infinite parallel plates is given by,

$E=\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_o}$

So, the field is independent of $\mathit{d}$.

Therefore, the net field can be calculated as:

$$\begin{gathered} E=E_1+E_2 \\ =\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_o}+\frac{\mathrm{σ}}{2{\mathrm{Î\mu }}_o} \\ =\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_o} \end{gathered}$$

Substitution in (1) yields,

$$\begin{gathered} E=\frac{47×{10}^{-9}}{8.85×{10}^{-12}} \\ =5.3×{10}^3\text{N/C} \end{gathered}$$

Thus, the magnitude of $\mathit{E}\mathbf{}$is $\mathbf{5}\mathbf{.}\mathbf{3}\mathbf{×}{\mathbf{10}}^{\mathbf{3}}\mathbf{}\mathbf{\text{N/C}}$.

(b)

Electric potential between parallel plates at any point is given by,

$V=Ed$

So, substitution in (2) yields,

$$\begin{gathered} V=2.2×{10}^{-2}×5.3×{10}^3 \\ =116.6\text{V} \end{gathered}$$

The potential difference between the two plates is$\mathbf{116}\mathbf{.}\mathbf{6}\mathbf{}\mathbf{\text{V}}$

(c)

If the distance between plates is doubled, it will make no changes to the field because the field is independent of d. For the voltage, according to (2), the voltage depends on d, so if d is doubled, then the potential between plates must be doubled.

Thus, E is dependent on d, so it won’t change, but the potential depends on d, so if d is doubled, then V must be doubled.