A capacitor consists of two parallel plates, each with an area of 17.0 cm2 , separated by a distance of 0.150 cm . The material that fills the volume between the plates has a dielectric constant of 4.00. The plates of the capacitor are connected to a 400 V battery.

A. What is the capacitance of the capacitor?

C = F
B. What is the charge on either plate?

Q = C
C. How much energy is stored in the charged capacitor?

U = J

Please click the following link so you can see, someone gave me a WRONG answer, so you dont repeat there same wrong steps or solutions:

a)   C = 4,012 10⁻¹⁴ F, b)  Q = 1.6 10⁻¹¹ C , c)   U = 3.21 10⁻¹¹ J

Explanation:

a) The capacitance of a capacitor is

       C = k e₀ A / d

Let's calculate

       C = 4 8.85 10⁻¹² 17 10⁻⁴ / 0.150 10⁻²

       C = 4,012 10⁻¹⁴ F

b) let's look  the charge

        C = Q / ΔV

         Q = C ΔV

         Q = 4,012 10⁻¹⁴ 400

         Q = 1.6 10⁻¹¹ C

c) The stored energy

        U = ½ C ΔV²

        U = ½ 4,012 10⁻¹⁴  400²

        U = 3.21 10⁻¹¹ J

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