Question

plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 85.0 nC . The plates are in vacuum. The electric field between the plates has a magnitude of 5.00×106 V/m . You may want to review (Pages 786 - 790) . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Properties of a parallel-plate capacitor. Part A What is the potential difference between the plates?

Answer 1

12500 V

Step-by-step explanation:

The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:

`\Delta V = E d`

where

`\Delta V` is the potential difference between the plates

E is the electric field strength

d is the distance between the plates

For the capacitor in this problem, we have

`E=5.00\cdot 10^6 V/m`

`d = 2.50 mm = 2.50\cdot 10^(-3) m`

Substituting, we find

`\Delta V = (5.00\cdot 10^6)(2.50\cdot 10^(-3))=12500 V`