A parallel-plate capacitor is connected across a potential difference of 10.0 V and stores 5.0 C of charge. The capacitor has a plate separation of 1.00 mm. What is the area of the plates?

Answer 1

The capacitance (C) of a capacitor is defined as the ratio of amount of charge (Q) to the electrical potential difference across the plates (V). In equation form: C = Q/V

The capacitance (C) of a parallel plate capacitor can also be expressed in terms of the relative permittivity of the material between the plates (ε), the distance separating the two plates (d) and the area of the faces of the plates (A). The equation is: C = εA/d

Now, this problem does not tell you that there is a dielectric material between the two plates, so the value of ε we will use is the permittivity of FREE space (ε_o) which is a fundamental constant of nature. So the equation becomes: C = ε_oA/d

The problem gives you the electrical potential difference V = 10.0 V, it gives you the charge Q = 5.0 C, and it gives you the separation distance between the plates d = 1.00 mm = 1.00×10^-3 m, and ε_o is a constant that we look up and find it is ε_o = 8.854×10^-12 F/m. Using the first equation and setting it equal to our most recent capacitance equation we find: Q/V = ε_oA/d

Now, using some algebra skills, we find that the area of the plate is: A = Qd/(ε_oV)

Plugging in our numbers: A = (5.0 C)(1.00×10^-3 m)/((8.854×10^-12 F/m)(10.0 V)) = 56.47165×10⁶ m²

So, the area of the plates is: A = 56×10⁶ m²