Slope: 5.651x10-14Use the equation U = Q2/(2C), to determine C0 using the slope of the graph and ompare this value of C0 with C0 in 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

Answered: Slope: 5.651x10-14 Use the equation U…

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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Question

Slope: 5.651x10^-14

Use the equation U = Q²/(2C), to determine C₀ using the slope of the graph and ompare this value of C₀ with C₀ in the table, then calculate the percentage error.

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?
E?
Stored energy
difference Q (C)
V (V)
Energy
between
U ()
(Volt)? | (C?)
(V/m)² |
Density u(J/m3)
u=U/(Ad)
plates E (V/m)
U= 1/2•€0•E4
*
1
-1.5
1.3x10-
1.0x10-
-.15
-2.25
1.69x1
-.0225
-9.9x10-14
13
13
0-26
2
-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-
2.0х10-
-.07
-0.49
3.6х10
-.0049
-2.2x10-14
14
14
-27
-0.4
4.0x10-
1.0x10-
-.04
-0.16
1.6x10| -.0016
-7.08x10-15
14
14
-27
.04
4.0x10-
1.0x10-
.04
0.16
1.6x10
.0016
7.08х10-15
14
14
-27
.07
6.0x10-
2.0x10-
.07
0.49
3.6х10
.0049
2.2x10-14
14
14
-27
8
1.0
9.0x10-
4.0x10-
.10
1.0
8.1.0x
.001
4.4x10-15
14
14
10-27
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

Expert Solution

Step 1

Using the slope value, calculate the the capacitance value $(C_{o})$ .

Formula used : $U = \frac{Q^{2}}{2 C}$

where, U is stored energy density $(i n J / m^{3})$ , Q is the charge $(i n C)$ and C is the capacitance $(i n F)$ .

$\Rightarrow \frac{Q^{2}}{U} = 2 C$

Plotting a graph of Square of the charge $(Q^{2})$ versus Stored energy density $(U)$ , the slope of the graph, should give the capacitance value (2 times the capacitance value).

$\Rightarrow \frac{Q^{2}}{U} = 5.651 x 10^{- 14} F$

(Note: Assuming that the units of slope is in fared (F).)

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Slope: 5.651x10-14 Use the equation U = Q2/(2C), to determine C0 using the slope of the graph and ompare this value of C0 with C0 in the table, then calculate the percentage error.