Slope: 4.414x10-14Use the equation U = 1/2 CV2 , to determine C0 using the slope, and compare this value of C0 with C0 above the table, then calculate the percentage error. Separation d (_10__ m) Plate Area A (__100m?) Capacitance C,=(ɛ,*A)/d__9.0x10^14_F)TrialPotentialChargeStoredElectric Fieldv2Q?E?Stored energydifference Q (C)V (V)EnergybetweenU ()(Volt)? | (C?)(V/m)² |Density u(J/m3)u=U/(Ad)plates E (V/m)U= 1/2•€0•E4*1-1.51.3x10-1.0x10--.15-2.251.69x1-.0225-9.9x10-1413130-262-1.31.2x10-7.0x10--.13-1.691.44x1-.0169-7.5x10-1413140-263-1.09.0x10-4.0x10--.10-1.08.1x10-.001-4.4x10-151414-274-0.76.0x10-2.0х10--.07-0.493.6х10-.0049-2.2x10-141414-27-0.44.0x10-1.0x10--.04-0.161.6x10| -.0016-7.08x10-151414-27.044.0x10-1.0x10-.040.161.6x10.00167.08х10-151414-27.076.0x10-2.0x10-.070.493.6х10.00492.2x10-141414-2781.09.0x10-4.0x10-.101.08.1.0x.0014.4x10-15141410-271.31.2x10-7.0x10-.131.691.44x1.01697.5x10-1413140-26101.51.3x10-1.0x10-.152.251.69x1.02259.9x10-1413130-26

The approach is mentioned in the following steps.

Separation d (_10_ m) Plate Area A (. 100 m:) Capacitance C.= (ɛ,*A)/d_(_9.0x10^14_F) Trial Potential Charge Stored Electric Field v2 Q? E2 Stored energy difference Q (C) V (V) (V/m)2 | Density uJ/m³) u=U/(Ad) Energy between (Volt)? (c2) U () plates E (V/m) U= 1/2•E0•E4 1 -1.5 1.3x10- 1.0x10- -.15 -2.25 1.69x1 -.0225 -9.9x10-14 13 13 0-26 -1.3 1.2x10- 7.0x10- -.13 -1.69 1.44x1 -.0169 -7.5x10-14 13 14 0-26 3 -1.0 9.0x10- 4.0x10- -.10 -1.0 8.1x10 -.001 -4.4x10-15 14 14 -27 4 -0.7 6.0x10- 2.0x10- -.07 -0.49 3.6x10 -.0049 -2.2x10-14 14 14 -27 5 -0.4 4.0x10- 1.0x10- -.04 -0.16 1.6x10 -.0016 -7.08x10-15 14 14 -27 6 .04 4.0x10- 1.0x10- .04 0.16 1.6x10 .0016 7.08x10-15 14 14 -27 7 .07 6.0x10- 2.0x10- .07 0.49 3.6x10 .0049 2.2x10-14 14 14 -27 1.0 9.0x10- 4.0x10- .10 1.0 8.1.0x .001 4.4x10-15 14 14 10-27 9 1.3 1.2x10- 7.0x10- .13 1.69 1.44x1 .0169 7.5x10-14 13 14 0-26 10 1.5 1.3x10- 1.0x10- .15 2.25 1.69x1 .0225 9.9x10-14 13 13 0-26 00

Separation d (_10_ m) Plate Area A (. 100 m:) Capacitance C.= (ɛ,*A)/d_(_9.0x10^14_F) Trial Potential Charge Stored Electric Field v2 Q? E2 Stored energy difference Q (C) V (V) (V/m)2 | Density uJ/m³) u=U/(Ad) Energy between (Volt)? (c2) U () plates E (V/m) U= 1/2•E0•E4 1 -1.5 1.3x10- 1.0x10- -.15 -2.25 1.69x1 -.0225 -9.9x10-14 13 13 0-26 -1.3 1.2x10- 7.0x10- -.13 -1.69 1.44x1 -.0169 -7.5x10-14 13 14 0-26 3 -1.0 9.0x10- 4.0x10- -.10 -1.0 8.1x10 -.001 -4.4x10-15 14 14 -27 4 -0.7 6.0x10- 2.0x10- -.07 -0.49 3.6x10 -.0049 -2.2x10-14 14 14 -27 5 -0.4 4.0x10- 1.0x10- -.04 -0.16 1.6x10 -.0016 -7.08x10-15 14 14 -27 6 .04 4.0x10- 1.0x10- .04 0.16 1.6x10 .0016 7.08x10-15 14 14 -27 7 .07 6.0x10- 2.0x10- .07 0.49 3.6x10 .0049 2.2x10-14 14 14 -27 1.0 9.0x10- 4.0x10- .10 1.0 8.1.0x .001 4.4x10-15 14 14 10-27 9 1.3 1.2x10- 7.0x10- .13 1.69 1.44x1 .0169 7.5x10-14 13 14 0-26 10 1.5 1.3x10- 1.0x10- .15 2.25 1.69x1 .0225 9.9x10-14 13 13 0-26 00

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

Chapter25: Electric Potential

Section: Chapter Questions

Problem 25.1OQ: In a certain region of space, the electric field is zero. From this fact, what can you conclude...

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