7. What is the potential difference across a parallel-plate capacitor whose plates are separated by adistance of 4.0 mm where each plate has a charge density of magnitude 5.0 pC/m?а.1.02 millivoltsb.1.43 millivoltsC.2.26 millivoltsd. 3.34 millivoltsе.4.43 millivolts Find the charge on capacitor, C2 , in the diagram below if Vab = 24.0 volts, C1 = 6.00 µF, C2 =3.00 µF, and C3 = 10.0 µF.12 μCb. 24 µC6.%3Dа.C.36 μCd. 48 μCе. 60 иСHEC1C2C3ab

7) Given,The charge density, σ=5 pC/m2we know, σ= QA⇒Q=σAwe also know,Q = CVC =…

Answered: Find the charge on capacitor, C2 , in…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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7. What is the potential difference across a parallel-plate capacitor whose plates are separated by a
distance of 4.0 mm where each plate has a charge density of magnitude 5.0 pC/m?
а.
1.02 millivolts
b.
1.43 millivolts
C.
2.26 millivolts
d. 3.34 millivolts
е.
4.43 millivolts

Find the charge on capacitor, C2 , in the diagram below if Vab = 24.0 volts, C1 = 6.00 µF, C2 =
3.00 µF, and C3 = 10.0 µF.
12 μC
b. 24 µC
6.
%3D
а.
C.
36 μC
d. 48 μC
е. 60 иС
HE
C1
C2
C3
a
b

Expert Solution

Step 1

$7) G i v e n, T h e c h a r g e d e n s i t y, σ = 5 p C / m^{2} w e k n o w, σ = \frac{Q}{A} \Rightarrow Q = σ A w e a l s o k n o w, Q = C V C = \frac{A ε_{0}}{d} Q = \frac{A ε_{0}}{d} V σ A = \frac{A ε_{0}}{d} V V = \frac{σ d}{ε_{0}} V = \frac{5 \times 10^{- 12} \times (4 \times 10^{- 3})}{(8.854 \times 10^{- 12})} V = 2.26 m i l l i v o l t s (o p t i o i n C i s c o r r e c t)$

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