Question

A parallel plate capacitor has a surface area of A = 46.1 cm2 and a plate separation of d = 10.7 mm. How much charge does the capacitor store, when it is connected to a battery supplying a voltage of V = 10.1 V?

Answer 1

38.4\times 10^{-12}

Step-by-step explanation:

A = Area of the capacitor plate = 46.1 cm² = 46.1 x 10⁻⁴ m²

d = separation between the plates of capacitor = 10.7 mm = 10.7 x 10⁻³ m

Capacitance of the capacitor is given as

`C = (\epsilon _(o)A)/(d)`

`C = ((8.85* 10^(-12))(46.1* 10^(-4)))/((10.7* 10^(-3)))`

`C = 3.8* 10^(-12)`

`V` = battery Voltage = 10.1 Volts

`Q` = Charge stored by the capacitor

Charge stored by the capacitor is given as

`Q = CV`

`Q = (3.8* 10^(-12))(10.1)`

`Q = 38.4* 10^(-12)`