voltage scale s

Question

Large voltages can be measured with a voltage balance. The electrical attraction that two charged capacitor plates exert on each other is compensated by a weight. The capacitor of the voltage scale has a capacity of C = 2.2 nF. The scale is balanced with no voltage applied. In this state, the distance between the capacitor plates is d = 1 cm. Now an additional weight of mass m = 33 g is added to the balance. In order to keep the distance between the plates constant at d = 1 cm, an initially unknown voltage U must now be applied between the capacitor plates. The force F, which generally acts between two capacitor plates, can be calculated using the following approach: F · d = WC . Here WC is the energy stored in the electric field of the capacitor. The gravitational acceleration is g = 9.81 m/s 2 , the electric field constant has the value ϵ0 = 8.8542 · 10−12 As

v m . Calculate the voltage U that is applied to the capacitor!

Answer 1

From your textbook,

WC = CU²/2.

Therefore

U = √(2WC/C)

= √(2Fd/C)

= √(2mgd/C).