91Ó°ÊÓ

The formula for calculating the electric field strength in a parallel capacitor is as follows:

$E=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}$

when the current is carried by one electrode $+\mathrm{σ}$surface charge density and the other one carries $-\mathrm{σ}$surface charge. In The field is 0 at first since the electrodes are neutral.. We can create $-\mathrm{σ}$on one electrode and $+\mathrm{σ}$in the other one by transferring electrons from one another. Each electron carries the charge of magnitude $e$. So by transferring $N$electrons we transfer the total charge of $Ne$and this creates the surface charge density of

$\mathrm{σ}=\frac{Ne}{a^2}$

where $a=4.0cm$. This yields

$E=\frac{Ne}{a^2{\mathrm{Î\mu }}_0}$

Solving for $N$we find

$N=\frac{a^2{\mathrm{Î\mu }}_{0}E}{e}$

Putting in $3.0.{10}^{6}N/c$we get

$N=2.7·{10}^{11}$

They need move $2.7.{10}^{11}$electrons by one electrode to the next.