Answered

An air-filled capacitor consists of two parallel plates, each with an area of 7.60 cm2, separated by a distance of 1.80 mm. If a 17.0 V potential difference is applied to these plates, calculate the following: ](a) the electric field between the plateskV/m(b) the capacitancepF(c) the charge on each platepC

Answer :

Khoso123

Answer:

(a) E=9.44×10³ V/m

(b) C=3.74pF

(c) Q=63.6pF

Explanation:

For Part (a)

The electric field between the plates is found by:

E=ΔV/d

Substitute  the given values

So:

[tex]E=\frac{17.0V}{1.80*10^{-3}m}\\ E=9.44*10^{3}V/m[/tex]

For Part (b)

The capacitance of parallel plate capacitor found from:

C=(∈₀A)/d

Substitute the given values

[tex]C=\frac{8.8542*10^{12}(7.60*10^{-4}m^{2} )}{1.8*10^{-3}m}\\ C=3.74*10^{-12}F\\ C=3.74pF[/tex]

For Part (c)

The Charge on each plate of capacitor is found by:

Q=CΔV

Substitute the  given values

[tex]Q=3.74*10^{-12}F*(17.0V)\\Q=63.6*10^{-12}C\\ Q=63.6pC[/tex]

Other Questions