91Ó°ÊÓ

Two $10-\mathrm{cm}$-diameter metal plates are suddenly connected together by a $0.224-\mathrm{mm}-$diameter copper wire stretched taut from the center of one plate to the center of the other.

According to Ohm's law: The maximal current is the ratio of the initial potential difference and the resistance of the wire,
$I_{\max }=\frac{U}{R}.$
The potential difference is seen in believing that the two plates form a capacitor, the charge is,
$C:=\frac{Q}{U}⇒U=\frac{Q}{C}$
The capacitance of an unfilled, parallel-plate capacitor will be,
$C={\mathrm{Î\mu }}_0\frac{A}{x},$
where $x$ indicates the distance. Then the potential difference will be,
$U=\frac{Q}{C}=\frac{Qx}{\mathrm{Î\mu }A}$

Covering that the area is that of a circle, allowing the diameter to be represented by $D$,$U=\frac{4Qx}{\mathrm{Î\mu }\mathrm{Ï€}{D}^2}.$
The resistance will be shown as,$R=\mathrm{ÒÏ}\frac{4x}{\mathrm{Ï€}{d}^2}$where$d$denotes the wire diameter .

To replace both terms for the potential difference and the resistance to the one for the current, obtaining
$I_{\max }=\frac{\frac{4Qx}{\mathrm{Î\mu }\mathrm{Ï€}{D}^2}}{\mathrm{ÒÏ}\frac{4x}{\mathrm{Ï€}{d}^2}},$
$I_{\max }=\frac{4Q{d}^2}{{\mathrm{Î\mu }}_0\mathrm{ÒÏ}{D}^2}$
$I_{\max }=\frac{4×1.25×{10}^{-8}×0.{000224}^2}{8.85×{10}^{-12}·1.7×{10}^{-8}0.1^2}$

$=1.7\mathrm{MA}$

The maximum current in the wire is$1.7MA$

The current $I$ is a discharge of electrical charge runners, which estimates the out pour of electric charge across the surface.

The current will decrease with time because the charge will have evolved lower. Thus, the potential difference will be more down reasonably.
So this decrease will be speedy.

Therefore, the current decrease with time.

The metal plates are suddenly connected together with the copper wire stretched taut from the center of one plate to the center of the other.

The considerable known expression for the energy accumulated in a capacitor is,

$E=\frac{C{U}^2}{2}$

Then substitute$U=Q/C$

To get$E=\frac{Q^2}{2C^2}$

To create one parametric substitution rather than two. Substituting to find,

$E=\frac{Q^2}{2\mathrm{Î\mu }x/\left(\mathrm{Ï€}{D}^2/4\right)}$

$=\frac{2Q^{2}x}{\mathrm{Ï€}\mathrm{Î\mu }{D}^2}$

$E=\frac{2×{\left(1.25×{10}^{-8}\right)}^2×0.01}{\mathrm{π}×8.85×{10}^{-12}×0.1^2}$

$E=11\mathrm{μ}\mathrm{J}$

The total amount of energy dissipated in the wire is $11\mathrm{μ}J$