- #1

012anonymousx

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## Homework Statement

What is the electric potential two-fifths of the way through a parallel-plate apparatus (from the positive plate) if the plates have a total separation of 5cm and field strength of 5000N/C.

## Homework Equations

Difference Potential Energy = Electric Field Strength x Distance

V = ed

## The Attempt at a Solution

= 5000 x (0.05 - ((2/5)(0.05))

= 5000 x (0.05-0.02)

= 5000 x (0.03)

= 150

Real answer is 250, so basically the whole distance of 0.05.

I thought that electric potential decreases as the current flows from positive to negative?

That is why, in my solution, I took the two-fifth of the whole distance away.

Edit: Please let me know if my solution as infact correct :)

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