91Ó°ÊÓ

a. The Potential of AA or AAA is 1.5 V. the difference between the positive terminal and negative terminal of a battery consider as the potential difference.

b.Within a parallel-plate capacitor, the direction of the electric filed goes from positive electrode to negative electrode.

The following equation shows the magnitude of the electric field strength ;

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

Here in role="math" localid="1648029333368" $\mathrm{η}consideredaschargedensity$

${\mathrm{Î\mu }}_{0}consideredasPermittivity$

The electric Potential inside a parallel-plate denoted as;

$V=Ed$; here d is the distance of two Plates.

The Potential Difference between the capacitors are as follows;

$$\begin{gathered} V=Ed \\ V=\left(\frac{\mathrm{η}}{{\mathrm{Î\mu }}_0}\right)d[As,E=(\frac{\mathrm{η}}{{\mathrm{Î\mu }}_0}\left)\right] \\ Or;V=\left(\frac{{\displaystyle \frac{Q}{A}}}{{\mathrm{Î\mu }}_0}\right)d[As,\mathrm{η}=(\frac{Q}{A}\left)\right] \\ Hence,Q=\frac{Vd}{A{\mathrm{Î\mu }}_0} \end{gathered}$$

The area of capacitor Plates;

$A=4cm×4cm=16cm=16×{10}^{-4}{m}^2$

The separation between the Plates;

$d=1.0mm=1.0×{10}^{-3}m$

The Eqiation of Supplied charge;

$Q=\frac{Vd}{A{\mathrm{Î\mu }}_0}$

Hence,$$\begin{gathered} Q=\frac{\left(1.6×{10}^{-4}\right)\left(8.85×{10}^{-12}{C}^2/N{m}^2\right)\left(1.5V\right)}{1.0×{10}^{-3}C} \\ ThereforetheSuppliedchargeis \\ Q=2.1×{10}^{-11}C \end{gathered}$$