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?

Respuesta :

Answer:

38.4\times 10^{-12}

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

[tex]C = \frac{\epsilon _{o}A}{d}[/tex]

[tex]C = \frac{(8.85\times 10^{-12})(46.1\times 10^{-4})}{(10.7\times 10^{-3})}[/tex]

[tex]C = 3.8\times 10^{-12}[/tex]

[tex]V [/tex] = battery Voltage = 10.1 Volts

[tex]Q [/tex] = Charge stored by the capacitor

Charge stored by the capacitor is given as

[tex]Q = CV [/tex]

[tex]Q = (3.8\times 10^{-12})(10.1) [/tex]

[tex]Q = 38.4\times 10^{-12} [/tex]