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SCIENCE · INVESTIGATE AND EXPLAIN

Integrate enclosed charge in a nonuniform sphere

A spherical Gaussian surface simplifies the field integral only when the complete charge distribution and boundary conditions have spherical symmetry. In this original insulating sphere, radial density is ρ0(1+αr²/R²), radius R=1 m and total charge Q=3 nC. No external field or conducting rearrangement is included.

Learning outcome: Compare α=0,1,2,3 at radii 0,0.5,1,2 m. Retain density normalization, enclosed charge and field separately. Identify which measurements distinguish density shape and which are identical because they depend only on total charge.

Starting recap

Review oriented normals, definite integrals, source volume measures and partial derivatives as one-coordinate rates. Potential differences use an explicit reference and field work uses a signed line integral.

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Try

For uniform density α=0, the enclosed charge at r=0.5 m is 3(0.5³)=0.375 nC and E=13.5 N/C. For α=2 with the same total Q, the enclosed charge there becomes 39/176≈0.221591 nC and E=351/44≈7.977273 N/C. Both models give 27 N/C at r=1 m and 6.75 N/C at r=2 m.

Open the experiment

Predict

Changing the radial density shape at fixed total charge leaves which field unchanged here?

Experiment

Compare α=0,1,2,3 at radii 0,0.5,1,2 m. Retain density normalization, enclosed charge and field separately. Identify which measurements distinguish density shape and which are identical because they depend only on total charge.

Normalize and integrate enclosed spherical charge

Worked teaching model · synthetic values · Insulating sphere radius1 m, fixed total charge3 nC and radial volume density rho(r)=rho0(1+αr²/R²) inside, zero outside. No external field. Exact spherical symmetry permits a radial Gaussian field. At center the continuous field is exactly zero; no division by zero is evaluated.

Reset keeps your prediction, captures and practice history.

Normalized volume-source field through its boundary

radial field (N/C)radius (m)02270

Unlike graph quantities rescale independently; read units and values, not screen angles. 1: piecewise Gaussian field (solid). Joined curves show the supplied model, not additional observations.

Equivalent numerical coordinates for Normalized volume-source field through its boundary

piecewise Gaussian field

radius (m)radial field (N/C)
00
0.0250.3070483
0.050.6154773
0.0750.9266676
0.11.242
0.1251.562855
0.151.890614
0.1752.226656
0.22.572364
0.2252.929116
0.253.298295
0.2753.681281
0.34.079455
0.3254.494196
0.354.926886
0.3755.378906
0.45.851636
0.4256.346457
0.456.86475
0.4757.407895
0.57.977273
0.5258.574264
0.559.20025
0.5759.856611
0.610.54473
0.62511.26598
0.6512.02175
0.67512.81342
0.713.64236
0.72514.50997
0.7515.41761
0.77516.36668
0.817.35855
0.82518.39459
0.8519.4762
0.87520.60476
0.921.78164
0.92523.00822
0.9524.28589
0.97525.61602
127
1.02525.69899
1.0524.4898
1.07523.36398
1.122.31405
1.12521.33333
1.1520.41588
1.17519.55636
1.218.75
1.22517.9925
1.2517.28
1.27516.609
1.315.97633
1.32515.37914
1.3514.81481
1.37514.28099
1.413.77551
1.42513.2964
1.4512.84185
1.47512.41023
1.512
1.52511.60978
1.5511.23829
1.57510.88435
1.610.54687
1.62510.22485
1.659.917355
1.6759.623524
1.79.342561
1.7259.073724
1.758.816327
1.7758.569728
1.88.333333
1.8258.106587
1.857.88897
1.8757.68
1.97.479224
1.9257.28622
1.957.100592
1.9756.921968
26.75

At most 82 representative computed coordinates including endpoints; the plotted curve can use a finer grid. A table of a revealed generator is not independent measurement evidence.

Calculated quantities under the declared measurement model
QuantityValueUnit
Density normalization0.3255442nC/m³
Enclosed charge0.2215909nC
Radial field7.977273N/C
Boundary field at radius 127N/C
Center field0N/C
Total sphere charge3nC

Normalize the full volume source, then integrate only the charge inside the selected Gaussian sphere.

Q=4πrho0 R³(1/3+α/5). For r≤R, Qenc=4πrho0[r³/3+αr⁵/(5R²)], and E=KQenc/r² by full spherical symmetry. At r=0 the limiting field is zero. No surface sheet is present, so the field is continuous at R even though volume density jumps; outside, only total Q enters. A thin shell with the same Q has a different interior field.

Predict before inspecting the values. Save contrasting records with assumptions and one alternative explanation. This virtual display does not certify practical measurement technique.

Normalize the full volume source, then integrate only the charge inside the selected Gaussian sphere.

Q=4πrho0 R³(1/3+α/5). For r≤R, Qenc=4πrho0[r³/3+αr⁵/(5R²)], and E=KQenc/r² by full spherical symmetry. At r=0 the limiting field is zero. No surface sheet is present, so the field is continuous at R even though volume density jumps; outside, only total Q enters. A thin shell with the same Q has a different interior field.

Captures preserve recipes and replay their calculated results. They are synthetic model evidence, not physical measurements.

Understand

Define the geometry and reference

A spherical Gaussian surface simplifies the field integral only when the complete charge distribution and boundary conditions have spherical symmetry. In this original insulating sphere, radial density is ρ0(1+αr²/R²), radius R=1 m and total charge Q=3 nC. No external field or conducting rearrangement is included.

Derive the integral or gradient

The volume element is 4πr²dr. Normalization gives Q=4πρ0R³(1/3+α/5). For an observer inside, integrate only to that radius: Qenc=4πρ0[r³/3+αr⁵/(5R²)]. Gauss then gives E=KQenc/r² outward. Outside, Qenc is the full Q. At r=0 use the limiting field zero rather than dividing by zero.

Check the inference boundary

At r=R the inside and outside fields match because there is no separate sheet of surface charge. The volume density may jump without making E jump. A uniformly charged thin shell of the same radius and total charge has the same exterior field but a different interior field; exterior data alone cannot distinguish those radial distributions.

Worked original case

For uniform density α=0, the enclosed charge at r=0.5 m is 3(0.5³)=0.375 nC and E=13.5 N/C. For α=2 with the same total Q, the enclosed charge there becomes 39/176≈0.221591 nC and E=351/44≈7.977273 N/C. Both models give 27 N/C at r=1 m and 6.75 N/C at r=2 m.

Investigation

Compare α=0,1,2,3 at radii 0,0.5,1,2 m. Retain density normalization, enclosed charge and field separately. Identify which measurements distinguish density shape and which are identical because they depend only on total charge.

Changed-case practice

Try the changed case before feedback

For the uniform α=0 sphere, find E at r=0.25 m in N/C.

This practice is public and unsaved. A matching number does not verify understanding. Retain your explanation, assumptions and changed-case evidence in the separate course workspace.

Sources and boundaries

Original calculations checked 30 October 2026 against the primary explanation. All source distributions, fields and readings are hypothetical. This is not a physical high-voltage or electrical handling activity, and public numerical feedback does not establish independent mastery. Preserve first predictions, critique, revision and newly assigned work in the separate retained course workspace.

Apply

Changing the radial density shape at fixed total charge leaves which field unchanged here?

Public practice checks a specific claim. It does not establish independent mastery.

Which geometry, calibration or source assumption could fail while an exact arithmetic fit remains possible?

Continue

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