Question

a. What is the potential of an ordinary AA or AAA battery? (If you’re not sure, find one and look at the label.)

b. An AA battery is connected to a parallel-plate capacitor having 4.0 cm * 4.0 cm plates spaced 1.0 mm apart. How much charge does the battery supply to each plate?

Short answer

a. Potential of AA or AAA is 1.5 V

b. The charge supplied by battery is$2.1\times {10}^{-11}C$

Step-by-step solution

Defination of potential 

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 ;

$E=\frac{\eta }{{\epsilon }_0}$

Here in role="math" localid="1648029333368" $\eta consideredaschargedensity$

${\epsilon }_{0}consideredasPermittivity$

The electric Potential inside a parallel-plate denoted as;

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

Equation For Charge within Capacitor

The Potential Difference between the capacitors are as follows;

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

Conversion of Units

The area of capacitor Plates;

$A=4cm\times 4cm=16cm=16\times {10}^{-4}{m}^2$

The separation between the Plates;

$d=1.0mm=1.0\times {10}^{-3}m$

Calculation of Supplied Charge

The Eqiation of Supplied charge;

$Q=\frac{Vd}{A{\epsilon }_0}$

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