The first Leyden jar was probably discovered by a German clerk named E. Georg von Kleist. Because von Kleist was not a scientist and did not keep good records, the credit for the discovery of the Leyden jar usually goes to physicist Pieter Musschenbroek from Leyden, Holland. Musschenbroek accidentally discovered the Leyden jar when he tried to charge a jar of water and shocked himself by touching the wire on the inside of the jar while holding the jar on the outside. He said that the shock was no ordinary shock and his body shook violently as though he had been hit by lightning. The energy from the jar that passed through his body was probably around 1 J, and his jar probably had a capacitance of about 1 nF. A) Estimate the charge that passed through Musschenbroek's body.

B) What was the potential difference between the inside and outside of the Leyden jar before Musschenbroek discharged it?

Question

Malia Gillespie

Physics

7 May, 15:59

The first Leyden jar was probably discovered by a German clerk named E. Georg von Kleist. Because von Kleist was not a scientist and did not keep good records, the credit for the discovery of the Leyden jar usually goes to physicist Pieter Musschenbroek from Leyden, Holland. Musschenbroek accidentally discovered the Leyden jar when he tried to charge a jar of water and shocked himself by touching the wire on the inside of the jar while holding the jar on the outside. He said that the shock was no ordinary shock and his body shook violently as though he had been hit by lightning. The energy from the jar that passed through his body was probably around 1 J, and his jar probably had a capacitance of about 1 nF. A) Estimate the charge that passed through Musschenbroek's body.

B) What was the potential difference between the inside and outside of the Leyden jar before Musschenbroek discharged it?

+2

Answers (1)

Know the Answer?

Dyer · Answer 1

a) q = 4.47 10⁻⁵ C

b) ΔV = 4.47 10⁴ V

Explanation:

A Leyden bottle works as a condenser that accumulates electrical charge, so we can use the formula of the energy stored in a capacitor

U = Q² / 2C

Q = √ (2UC)

let's reduce the magnitudes to the SI system

c = 1 nF = 1 10⁻⁹ F

let's calculate

q = √ (2 1 10⁻⁹-9)

q = 0.447 10⁻⁴ C

q = 4.47 10⁻⁵ C

b) for the potential difference we use

C = Q / ΔV

ΔV = Q / C

ΔV = 4.47 10⁻⁵ / 1 10⁻⁹

ΔV = 4.47 10⁴ V