Two Hat conductors are placed with their in- ner faces separated by 16 mm. If the surface charge density on inner face A is 105 pC/m and on inner face B is -105 pC/m², calculate the electric po- tential difference AV VA – VB- Use 8.85419 x 10¬2 c² /Nm2. Answer in units of V. %3D

Two Hat conductors are placed with their in- ner faces separated by 16 mm. If the surface charge density on inner face A is 105 pC/m and on inner face B is -105 pC/m², calculate the electric po- tential difference AV VA – VB- Use 8.85419 x 10¬2 c² /Nm2. Answer in units of V. %3D

Physics for Scientists and Engineers, Technology Update (No access codes included)

9th Edition

ISBN:9781305116399

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter24: Gauss’s Law

Section: Chapter Questions

Problem 24.46P: A thin, square, conducting plate 50.0 cm on a side lies in the xy plane. A total charge of 4.00 108...

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Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

Question

Two flat conductors are placed with their in-
ner faces separated by 16 mm.
If the surface charge density on inner
face A is 105 pC/m and on inner face B
is -105 pC/m², calculate the electric po-
tential difference AV
VA
VB-
Use
€0 = 8.85419 x 10¬12 c² /Nm².
Answer in units of V.
%3D

Expert Solution

Step 1

Given

Separation between the two plates is $d = 16 m m = 16 \times 10^{- 3} m$

Magnitude of charge density of the two plates is $σ = 105 p C / m^{2} = 105 \times 10^{- 12} C / m^{2}$

We have to calculate the potential difference between the two plates.

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