A capacitor is made from two 1.2 cm diameter coins separated by a 0.18 mm thick piece of paper (K = 3.7). A 6 V battery is connected to the capacitor. How much charge is on each coin?

My solution:
C= K * Permitivity * A/d = q/v

I solved for q and got 1.2 X 10^-10

But i still got it wrong. Can someone please check and see where i went wrong or check to see if you arrive at a different answer

Area = 1.131*10^-4 m^2

d = 1.8*10^-4 m
epsilon = 8.85*10^-12
C = 2.06*10^-11 F
q = CV = 1.2*10^-10 coulombs

I agree with your answer

To calculate the charge on each coin in the given capacitor, we can use the equation q = CV, where q is the charge, C is the capacitance, and V is the voltage.

To find the capacitance, we can use the equation C = (K * ε * A) / d, where C is the capacitance, K is the dielectric constant, ε is the permittivity of free space (8.85 x 10^-12 F/m), A is the area of the coin, and d is the separation between the coins.

First, let's calculate the area of the coin. The diameter is given as 1.2 cm, so the radius (r) is half of that, which is equal to 0.6 cm or 0.006 m. The area (A) of the coin is then given by A = π * r^2.

A = π * (0.006)^2 ≈ 0.000113097 m^2

Now we can substitute the values into the capacitance equation:

C = (K * ε * A) / d

C = (3.7 * 8.85 x 10^-12 F/m * 0.000113097 m^2) / (0.00018 m)

C ≈ 2.306 x 10^-11 F

Now we can find the charge on each coin using the equation q = CV:

q = (2.306 x 10^-11 F) * (6 V)

q = 1.384 x 10^-10 C

Therefore, the charge on each coin is approximately 1.384 x 10^-10 Coulombs.