The plates of an empty parallel-plate capacitor of capacitance 5.0 picofarads are 2.0 millimetres apart. What is the area of each plate?
We can call the capacitance of 5.0 picofarads 𝐶 and the distance separating the parallel plates 2.0 millimetres 𝑑. We want to know the area of each plate in the capacitor. We’ll call that area 𝐴.
We can begin our solution by recalling the mathematical relationship for the capacitance of a parallel-plate capacitor. That capacitance 𝐶 is equal to the permittivity of free space ε nought times the area of the plates over the distance separating them. ε nought is a constant, whose value we’ll assume is exactly 8.85 times 10 to the negative 12th farads per metre.
Rearranging this equation to solve for the area 𝐴, we find it’s equal to 𝐶 times 𝑑 over ε nought. We’re given both 𝐶 and 𝑑 in the problem statement and ε nought is a known constant. We are now ready to plug in and solve for 𝐴.
When we do, we’re careful to use units of farads for our capacitance and metres for our distance, converting them from the units originally given. Entering these values on our calculator, we find that 𝐴 to two significant figures is 1.1 times 10 to the negative third metre squared. That’s the area of each plate of this parallel-plate capacitor.