Question

Two 7.0cm×7.0cm metal electrodes are spaced 1.0 mm apart and connected by wires to the terminals of a 9.0 v battery.

Answer 1

`q=390.285x10^(-12)C`

Step-by-step explanation:

Those kind of problems of electric physics is about capacitors, so the normal questions are:

What are the charge on each electrode?

Solve this you can get other information required in the problem or write down the other questions you need

`7.0cm*(1m)/(100cm)=70x10^(-3)m`

`A=70x10^(-3)m*70x10^(-3)m=4.9x10^(-3)m62`

Capacitance

`C=E_o*(A)/(d)`

`E_o=8.85x10^(-12)(C^2)/(N*m^2)`

`C=8.85x10^(-12)(C^2)/(N*m^2)*(4.9x10^(-3)m^2)/(1x10^(-3)m)`

`C=43.365x10^(-12)F`

`C=43.365pF`

The charge is find by the equation

`q=C*V`

`q=43.365pF*9V`

`q=390.285x10^(-12)C`