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?

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:

where

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

Substituting, we find

f=ma newton

f=100x2.5=250n

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