   Chapter 16, Problem 58AP

Chapter
Section
Textbook Problem

When a potential difference of 150. V is applied to the plates of an air-filled parallel-plate capacitor, the plates carry a surface charge density of 3.00 × 10−10 C/cm2. What is the spacing between the plates?

To determine
The spacing between the plates.

Explanation

Given info: The surface charge density is ( σ ) is 3.00×1010C/m2 . The potential difference ( ΔV ) is 150 V.

Formula to calculate the charge is,

Q=σA

Alternatively, the charge is given by,

Q=ε0A(ΔV)d

Equating the equations,

ε0A(ΔV)d=σA

On Re-arranging,

d=ε0(ΔV)σ

Substitute 8.85×1012m3kg1s4A2 for ε0 , 150 V for ΔV and 3

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