# Problem: The electric potential inside a parallel-plate capacitor __________.decreases inversely with the square of the distance from the negative platedecreases linearly from the negative to the positive plateincreases linearly from the negative to the positive plateis constantdecreases inversely with distance from the negative plate

###### FREE Expert Solution

We are required to explain the change in the electric potential in a parallel plate capacitor.

Electric field:

$\overline{){\mathbf{E}}{\mathbf{=}}{\mathbf{-}}\frac{\mathbf{∆}\mathbf{V}}{\mathbf{∆}\mathbf{x}}}$

The electric field in parallel plate capacitors:

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{\sigma }}{{\mathbf{\epsilon }}_{\mathbf{0}}}}$ σ is the magnitude of the charge on the plates

94% (129 ratings) ###### Problem Details

The electric potential inside a parallel-plate capacitor __________.

 decreases inversely with the square of the distance from the negative plate decreases linearly from the negative to the positive plate increases linearly from the negative to the positive plate is constant decreases inversely with distance from the negative plate