View-101210 13-10-13.Plate ChargesElectric Field LinesMeters-СарacitancePlate ChargeStored EnergyDisconnect Battery|VoltmeterCapacitancePlate Charge (top)Stored EnergyO Electric Field Detecto0.71x10-12 F4.01x1013 c1.13x10-13 JReset ASeparation5.0 mm0.565 V1.5 VPlate Area400.0 mm?OV-1.5 VO-lo- IC, =4, using e,= 8.85x1012_ dN-mTable 1.• Separation, d =• Plate Area, A =• Capacitance, C,mm?PotentialStoreddifferenceChargeEnergyQTrialAVQ|(0.1–1.5 V)(V)(C)(J)(volt)(C')1

Separation, d = 5 mm = 0.005 m Potential difference, V = 0.565 V Area of plate, A = 400 m2 =…

Answered: Table 1. • Separation, d=, • Plate…

Table 1. • Separation, d=, • Plate Area, A =, • Capacitance, c, = = m? %3! Potential Stored difference Charge Energy Q? Trial AV Q U (0.1-1.5 V) (V) (C) (J) (volt)? (C')

College Physics

1st Edition

ISBN:9781938168000

Author:Paul Peter Urone, Roger Hinrichs

Publisher:Paul Peter Urone, Roger Hinrichs

Chapter19: Electric Potential And Electric Field

Section: Chapter Questions

Problem 14PE: What is the strength of the electric field between two parallel conducting plates separated by 1.00...

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Question

View-
1012
10 13-
10-13.
Plate Charges
Electric Field Lines
Meters-
Сарacitance
Plate Charge
Stored Energy
Disconnect Battery
|Voltmeter
Capacitance
Plate Charge (top)
Stored Energy
O Electric Field Detecto
0.71x10-12 F
4.01x1013 c
1.13x10-13 J
Reset A
Separation
5.0 mm
0.565 V
1.5 V
Plate Area
400.0 mm?
OV
-1.5 V
O-lo-

IC, =4, using e,
= 8.85x1012_ d
N-m
Table 1.
• Separation, d =
• Plate Area, A =
• Capacitance, C,
m
m?
Potential
Stored
difference
Charge
Energy
Q
Trial
AV
Q
|(0.1–1.5 V)
(V)
(C)
(J)
(volt)
(C')
1

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