Answer:
The amount of charge is [tex]q= 8.342 *10^{-14}C[/tex]
Explanation:
Here let estimate that the battery is a parallel whose plates are of area = [tex]\pi r^2[/tex] substituting value
[tex]= 3.142*\frac{1}{100} = 3.142 *10^{-4}m^2[/tex]
And the this plates are separated by a length d= 5.0cm
Now looking at this estimate we can see that the setup looks like a capacitor
So to calculate the capacitance of our estimated capacitor we have
[tex]C = \frac{\epsilon A}{d}[/tex]
Where [tex]\epsilon[/tex] is the permitivity of free space with value [tex]= 8.85*10^{-12}\ C^2/Nm^2[/tex]
substituting values
[tex]C = \frac{8.85*10^{-12} * 3.142 *10^{-4}}{0.050}[/tex]
[tex]=5.56 *10^{-14} F[/tex]
The charge is mathematically represented as
[tex]q = CV[/tex]
Substituting values
[tex]q =5.56*10^{-14} *1.5[/tex]
[tex]= 8.342 *10^{-14}C[/tex]
