Question

A parallel-plate capacitor with plate area 2.5 cm^2 and air-gap separation 0.50mm is connected to a 6.0-V battery, and fully charged. The battery is then disconnected. What is the charge on the capacitor?

Answer 1

The charge on the capacitor is 5.31 x 10^-10 C.

Step-by-step explanation:

To calculate the charge on the capacitor, we can use the equation Q = CV, where Q is the charge, C is the capacitance, and V is the voltage. In this case, the voltage is 6.0V and the capacitance can be calculated using the formula C = ε₀A/d, where ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between the plates. The permittivity of free space is approximately 8.85 x 10^-12 F/m.

Converting the units, the plate area is 2.5 cm², which is equal to 2.5 x 10^-4 m², and the separation is 0.50mm, which is equal to 0.50 x 10^-3 m. Plugging the values into the equation, we have C = (8.85 x 10^-12 F/m)(2.5 x 10^-4 m²) / (0.50 x 10^-3 m), which gives us a capacitance of 8.85 x 10^-11 F.

Now we can calculate the charge using the equation Q = CV. Substituting the values, we get Q = (8.85 x 10^-11 F)(6.0V), which gives us a charge of 5.31 x 10^-10 C.