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

Answer:

12500 V

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

$\Delta V = (5.00\cdot 10^6)(2.50\cdot 10^{-3})=12500 V$