How To Find Electric Flux Through A Sphere

Flux, Gauss' police force

Flux

Problem:

A deejay with radius r = 0.x 1000 is oriented with its normal unit vector n at an angle of thirty^o to a uniform electric field E with magnitude ii.0*ten^three Due north/C.
(a)  What is the electric flux through the disk?
(b)  What is the flux through the deejay if it is turned so that its normal is perpendicular to Eastward?
(c)  What is the flux through the deejay if its normal is parallel to East?

Solution:

• Concepts:
  Electric flux
• Reasoning:
  The flux is given past Φ = ∫Due east∙dA.
• Details of the calculation:
  (a)  Since the field is constant, Φ = EAcosθ, where θ is the bending betwixt the field and the normal, and A = πr^two = π(0.1)ⁱⁱ = 0.0314 m.  Then for θ = 30^o nosotros obtain Φ = 54.four Nm²/C.
  (b)  Now θ = xc^o thus Φ = 0 Nmⁱⁱ/C.
  (c)  Now θ = 0^o thus Φ = 63 Due north grand²/C.

---

Gauss' law, spherical symmetry

Problem:

A solid conducting sphere of radius 2 cm has a accuse of 8 microCoulomb.  A conducting spherical shell of inner radius 4 cm and outer radius 5 cm is concentric with the solid sphere and has a charge of -4 microCoulomb.
(a)  What is the magnitude and direction of the electric field at r = 1 cm?
(b)  What is the magnitude and direction of the electrical field at r = three cm?
(c)  What is the magnitude and direction of the electric field at r = 4.5 cm?
(d)  What is the magnitude and direction of the electric field at r = 7 cm?

Like problem:

Problem:

Inside a sphere of radius R and uniformly charged with the volume charge density ρ, there is a neutral spherical cavity of radius R₁ with its center a altitude a from the center of the charged sphere.  If (R₁ + a) < R, detect the electric field within the cavity.

Solution:

• Concepts:
  Gauss' police force, the principle of superposition
• Reasoning:
  Nosotros tin can view the sphere with the crenel every bit a superposition of a sphere with radius R having a compatible charge density ρ and another sphere with radius R₁ located in the cavity infinite having a compatible charge density -ρ.  The field due to each of these spherical charge distributions tin can be found from Gauss' police force.
• Details of the calculation:
  Permit the heart of the large sphere be located at the origin.
  East = E ₁ + Due east ₂.
  E ₁ = ρr/3ε₀  for r < R.
  E ₂ = -ρ(r-a)/3ε₀  within the crenel.
  Inside the cavity nosotros therefore have E = ρr/3ε₀ - ρr/3ε₀ + ρa/3ε₀ = ρa/3ε₀
  The field within the cavity is abiding and points into the direction of the vector a.

Problem:

The electrostatic potential Five is 0 on a spherical shell of radius r_ane = 0.10 m and V is 100 Volts for a concentric shell of radius r₂ = 2.00 m.  Assume free space between these concentric shells, and find E between the shells.

Solution:

• Concepts:
  Gauss' law
• Reasoning:
  The problem has enough symmetry to find E(r) from Gauss' law alone.
• Details of the calculation:
  From Gauss' law: Due east(r ) ∝ 1/rⁱⁱ between the spheres.
  The direction of E is radially in.  E(r) = -A/r², wit A positive.
  And then 5₀ = -∫_r1 ^r2East(r)dr = A∫_r1 ^r2(one/r²)dr = -A(ane/r₂ - 1/r₁) - 100 V
  A(one/(0.1 g) - i/(2m)) = 100 V.
  A = (100/9.5)  Vm.
  E = -10.526 Five/k (r/r).

Problem:

A charge distribution produces an electric field E = A(1 - exp(-βr))(r/r³) where A and β are constants.  Find the cyberspace accuse inside the radius r = i/β.

Solution:

• Concepts:
  Gauss' law
• Reasoning:
  A radial field is produced by a spherically symmetric charge distribution.
• Details of the calculation:
  Due east(r)4πrⁱⁱ = Q_within/ε₀ in SI units.
  For r = 1/β we have A(1 - e^-1)4π = Q_within/ε₀.
  Q_inside = A(1 - e^-1)4πε₀.

Problem:

Determine the charge distribution that volition give ascension to the potential V(r) = kq exp(-mr)/r, with m a positive constants.  Summate the total charge in the distribution.

Solution:

• Concepts:
  Gauss' police force, ∇ⁱⁱV(r) = -ρ/ε₀,
• Reasoning:
  V(r) = V(r).  ∇²V(r) = -ρ/ε₀, except at r = 0, where the expression for ∇^twoV(r) = is non defined.
  To discover the charge at the origin we utilise Gauss' law.
• Details of the adding:
  ∇²5(r) = (1/r²)(∂/∂r)(rⁱⁱ∂V(r)/∂r) = m² kq exp(-mr)/r = 1000²V(r) = -ρ/ε₀,
  except at r = 0, where the above expression for ∇^two5(r) = is not defined.
  To find the charge at the origin we use Gauss' law.
  E = -∇V(r) = (∂V(r)/∂r)(r/r).  E_r(r) = (kq/r²)exp(-mr) + kq m exp(-mr)/r.
  For a spherical surface of radius r we have 4πr²E(r) = Q_within/ε₀.
  As r --> 0, Q_inside/ε₀ = 4π kq.
  The charge distribution that gives rise to the potential V(r) = kq exp(-mr)/r therefore is
  ρ(r) = 4πε₀ kqδ(r) - ε₀m² kq exp(-mr)/r.
  With grand = one/(4πε₀ ) we have ρ(r) = qδ(r) - (thou²/4π)q exp(-mr)/r.
  The total charge in the distribution is given by
  Q = ∫_{all space} ρ(r)dV = q∫_{all infinite} δ(r)dV - (m²/4π)q 4π ∫₀ ^∞ rⁱⁱ exp(-mr)/r dr
  = q - m²q/m² = 0.

Trouble:

(a)  Consider a non-conducting sphere with radius a.  This sphere carries a cyberspace accuse Q, assumed to be uniformly distributed.  Find the electric field inside and outside the sphere.  Sketch the result.
(b)  Now consider a conducting sphere with radius a carrying a net charge Q.  Find the electric field inside and outside the sphere.

Solution:

• Concepts:
  Gauss' law, Φ_e = ∫_{airtight surface} E·dA = Q_inside/ε₀
• Reasoning:
  The charge distribution has spherical symmetry and Eastward can be found from Gauss' police force alone.
• Details of the calculation:
  (a)  Place the center of the sphere at the origin of the coordinate system.  Consider a spherical Gaussian surface of radius r centered at the centre of the spherical accuse distribution.
  i.   Let r be greater than a, so that the surface encloses the entire charge distribution.
  The electric field is radial, the vector E is normal to any surface chemical element dA.   Thus flux through the surface is  Φ_e = ∫ E·dA = ∫EdA = E 4πr^two = Q_within/ε₀ = Q/ε₀
  E = Q/(4πε₀r²) north, where n = r/r.
  The field outside the sphere looks like the field of a point charge Q.
  ii.  Let r be smaller than a, then that the surface only encloses a function of the accuse distribution.  At present Q_inside is the charge density ρ = Q/V times the volume 4πr³/iii of the distribution which lies inside the spherical Gaussian surface.  We therefore accept
  E = ρr/(3ε₀) north = Qr/(4πε₀a³) north.
  The field inside the sphere increases linearly with r.
  
  (b)  The accuse is uniformly distributed over the surface of the conductor.
  r >a: Due east = Q/(4πε₀rⁱⁱ) n,  r < a: E = 0, the electric field inside a conductor is cipher in electrostatics.

Problem:

A sphere of radius R has volume charge density ρ = Krⁿ, for some constants K and n. The region r > R is filled with a usher (all the way to infinity).
(a)  Discover the volume accuse density ρ in the region r > R, inside the usher, and the surface charge density at r = R.
(b)  Find the electric field E everywhere, i.e. for r < R and for r > R.
(c)  Find the potential Φ everywhere, taking Φ to vanish at infinity.
(d)  How much energy is stored in this system?

• Concepts:
  Gauss' law, properties of conductors
• Reasoning:
  The problem has spherical symmetry.  Within a conductor ρ and E are nix.
• Details of the calculation:
  (a)  ρ = 0 in a usher, and since E = 0, the total charge enclosed by a Gaussian surface in a conductor is zip, then the surface charge at r = R cancels that of the charged sphere.
  σ = (1/(4πR²))∫₀ ⁰Kr^north 4πrⁱⁱ dr = -KR^n+one/(n + 3)
  The surface charge is uniformly distributed on the inner surface of the conductor.
  (b)  By symmetry Eastward(r) = East(r)r/r.  Gauss' law practical to a Gaussian spherical surface of radius r < R gives
  E(r)4πr^two = -4πKrⁿ⁺³/((n + 3)ε₀),  Eastward(r) = -Krⁿ⁺ⁱ/((n + iii)ε₀),  for r < R.
  Once more E = 0 for r > R, inside the usher.
  (c) Writing the above Due east every bit -∇Φ, we obtain
  Φ = - K(r^north+2 -  R^northward+2)/((northward + 2)(due north + iii)ε₀) for r < 0,  Φ = 0 for r > 0.
  (d)  U = ½∫ρΦdV = (ε₀/2)∫E²dV = 2πK^twoR²ⁿ⁺⁵)/((2n + 5)(n + three)²ε₀)

Problem:

Ii metallic spheres of the same radius r are immersed in a homogeneous liquid with resistivity ρ.  What is the total resistance between two spheres?  Presume that the distance between ii spheres is much larger than the sphere radius.

Solution:

• Concepts:
  Gauss' law, definition of I, R, and C
• Reasoning:
  Consider ii conductors carrying free charges +Q and -Q respectively.
  Consider a closed surface S surrounding the conductor carrying charge Q.
  The simply excess charges inside S reside on the conductor.

  From Gauss' law:               Q = ε₀∫_S E∙due north dS
  From the definition of I:     I = ∫_S j∙n dS

• Details of the adding:
  Using j = σE we have Q/I = RQ/V = RC = ε₀/σ.
  If the distance between two spheres is much larger than the radius of each sphere, and then we take V = [i/(4πε₀)](Q/a) - (-Q/a)) = [i/(2πε₀)](Q/a)
  C = Q/V = 2πε₀a.  Therefore  R = ane/(σ2πa) =  ρ/(2πa).

Problem:

A charge Q is uniformly distributed through the volume of a sphere of radius R.  Calculate the electrostatic energy stored in the resulting electrical field.Solution:

• Concepts:
  Gauss' constabulary, electrostatic energy
• Reasoning:
  A radial field is produced by a spherically symmetric charge distribution.
• Details of the calculation:
  From Gauss' law we have E(r) = Q/(4πε₀r²) for r > R, and East(r) = ρr/(3ε₀) for r < R.
  Here ρ = 3Q/4πR³.  Therefore Due east(r) = Qr/(4πRⁱⁱⁱε₀) for r < a.   At r = 0 we accept E(r) = 0.
  The total electrostatic free energy of the sphere may be calculated from the electric field, using U = (ε₀/2) ∫_{all space} E·E dV.   (SI units).
  U = (4πε₀/2)[∫₀ ^Rr²dr Q²r^two/(4πR^threeε₀)² + ∫_R ^∞r²dr Q²/(4πε₀r²)²]
  = (Qⁱⁱ/(8πε₀))[∫₀ ^Rr^ivdr/R⁶ + ∫_R ^∞dr /r²] = [Q^two/(8πε₀)][1/(5R) + 1/R] = (3/5)Q²/(4πε₀R).

Problem:

A model of the hydrogen cantlet was proposed before the advent of quantum mechanics, which consists of a single electron of mass m and an immobile uniform spherical distribution of positive accuse with radius R.  Assume that the positive charge interacts with the electron via the usual Coulomb interaction only otherwise does not offer any resistance to the move of the electron.
(a)  Explain why the electron's equilibrium position is at the center of the positive accuse.
(b)  Evidence that the electron will undergo simple harmonic motion if it is displaced a distance d < R away from the centre of the positive accuse.  Calculate its frequency of oscillation.
(c)  How big would the atom need to exist in lodge to emit red light with a frequency of iv.57*ten¹⁴ Hz?  Compare your answer with the radius of the hydrogen atom.
(d)  If the electron is displaced a distance d > R from the center, will it oscillate in position?  Will it undergo uncomplicated harmonic motion?  Explain!

Solution:

• Concepts:
  Gauss' law
• Reasoning:
  The field due to a spherically symmetric charge distribution can be found from Gauss' law.
• Details of the calculation:
  The charge distribution is spherically symmetric and and so the electric field is E(r) = Eastward(r) r/r.
  Let the eye of the positive charge exist the origin and consider a Gaussian surface defined to be a sphere or radius r at the same origin.
  The total flux is Φ = ∫E·dA = 4πrⁱⁱE(r) = Q_inside/ε₀.
  If 0 < r < R, then Q_inside = 4πρr³/3, with ρ = 3q_e/4πr³.
  In this region E(r) = ρr/(3ε₀).
  If r > R, the electric field is the same every bit expected for a point charge, Due east(r) = q_e/4π ε₀r².
  We can now easily respond the questions.
  (a)  The force acting on the electron is F(r) = -q_eρr/(3ε₀), pointing towards the origin.  Therefore the only point where the sum of all forces is nil is the origin.
  (b)  The force is a restoring strength and is proportional to the displacement of the electron from r = 0.  The resulting motion will be simple harmonic motion.
  The "spring constant" associated with this movement is grand = q_{due east}ρ/(3ε₀) = q_east ⁱⁱ/(4πε₀R³).
  The frequency of the oscillations is therefore
  f = (2π)^-i(k/yard)^1/2 = (2π)^-1(q_e ²/(4πε₀mR³))^one/2.
  (c)  R = [q_east ²/(16π³ε₀mf²)]^1/3.  For f = 4.57*1014 Hz we demand R = 3.13*10^-10 m.
  (d)  For d > R the force is still restoring.  Therefore the electron will undergo oscillatory motion.  The magnitude of the strength is proportional to 1/r², however, and therefore the motion will not be uncomplicated harmonic motility.

Problem:

A spherical charge distribution is given by
ρ = ρ₀(1 - r/a),    r < a,
ρ = 0,    r > a.
(a)  Summate the total accuse Q.
(b)  Find the electric field and potential for r > a.
(c)  Find the electrical field and potential for r < a.
(d)  Find the electrostatic energy of this accuse distribution.

Solution:

• Concepts:
  Gauss' police
• Reasoning:
  The field due to a spherically symmetric charge distribution tin can be found from Gauss' constabulary.
• How do they employ?
  (a)  Q = 4πρ₀∫₀ ^a r^twodr(1 - r/a) = 4πρ₀[(a^three/three) - (a³/4)] = 4πρ₀a³/12.
  (b)  E = (one/(4πε₀))Q/r^two, radially outward for positive ρ₀.
  Due east = (i/ε₀)ρ₀a³/(12r^two).
  Φ = (1/(4πε₀))Q/r = (1/ε₀)ρ₀a^three/(12r).
  (c)  E = (1/(4πε₀)) Q_inside/r², radially outward for positive ρ₀.
  Q_within = Q = 4πρ₀∫₀ ^r r'²dr'(i  -r'/a)  = 4πρ₀[(r³/3) - (r⁴/(4a))].
  East = (i/ε₀)ρ₀[(r/3)  -(r²/(4a))].
  Φ = Φ(a) + ∫_r ^aE(r)dr = Φ(a) + (ρ₀/ε₀)∫_r ^a[(r/3) - (r^two/(4a))]dr
  = Φ(a) + (ρ₀/ε₀)[(r^two/6) - (rⁱⁱⁱ/(12a))]_r ^a
  = (ρ₀/ε₀)[a²/12 + a²/6 - a^two/12  - r²/6 + r³/(12a)]
  = (ρ₀/ε₀)[a²/6  - r^two/half-dozen + r^three/(12a)].
  (d) U = (ε₀/ii)∫_{all_space} E·East dV
  U = (ρ₀ ²/2ε₀)∫₀ ^a[(rⁱⁱ/9) - (r³/(6a)) + (r^iv/(16aⁱⁱ))]4πrⁱⁱdr
  + (ρ₀ ⁱⁱa^{half dozen}/288ε₀)∫_a ^∞(i/r⁴)4πr²dr
  = (ρ₀ ²a⁵/ε₀)*0.0648.
  or
  U = (1/2)∫₀ ^aρ(r)Φ(r)dV = (4π/2)∫_r ^aρ(r)Φ(r)rⁱⁱdr
  = (2πρ₀ ²/ε₀)∫₀ ^a(i - r/a)[a^two/6  - rⁱⁱ/6 + r^three/(12a)]r^twodr
  = (ρ₀ ⁱⁱa^five/ε₀)*0.0648.

Similar trouble

---

Gauss' police, cylindrical symmetry

Problem:

A 500 g length of high-voltage cable is undergoing electrical testing.  The cablevision consists of ii coaxial conductors, the inner of five mm bore and the outer of 9 mm internal diameter.  The space between the conductors is filled with polythene which has a relative permittivity of 2 and which can withstand electric field strength of 60 MVm^-1.
(a)  Discover the maximum voltage which can be applied between the conductors and the free energy stored in the cable at this voltage.
(b)  If the cablevision is to exist discharged to a safe level of 50 V in i minute, what value of resistance must exist continued across it?  What is the maximum ability and the total free energy dissipated in the resistance?

Solution:

• Concepts:
  Gauss' law, the cylindrical capacitor
• Reasoning:
  The problem has plenty symmetry to find D(r) and Eastward(r) from Gauss' police force alone.
• Details of the calculation:
  (a)  From Gauss' police: D(r), Due east(r ) ∝  i/r betwixt the cylinders.
  Let the outer cylinder be grounded and Five₀ be positive.  Then the direction of E is radially outward.
  E(r) = A/r, with A positive.  Then V₀ = -∫_b/2 ^a/2Eastward(r)dr = A∫_a/two ^b/2(1/r)dr
  = A(lnb/two - lna/2) = Aln(b/a).  A = 5₀/ln(b/a),
  with a = 5 mm and b = 9 mm.
  E(r) = V₀/(rln(b/a)).
  If Eastward_max = E(a/2) = 2V₀/(aln(b/a)) = lx MV/thou, then
  V₀ = (half dozen*10⁷ V/m)*(ii.five*10^-3 k)*ln(9/5) = 8.82*10⁴ V is the maximum voltage which can be applied.
  Free energy stored in the cable:
  Due west = (ε/2)∫Due eastⁱⁱdV = (εL/2)2π∫_a/2 ^b/2Eⁱⁱrdr = επL∫_a/2 ^b/ii(V₀/(rln(b/a)))²rdr.
  W = [επLV₀ ²/ln(b/a)] = [2πε₀(500 chiliad)*(8.82*ten⁴ Five)ⁱⁱ/ln(9/five)] = 368 J.

Problem:

A plasma is generated within a long hollow cylinder of radius R.  It has the charge distribution
ρ(r) = ρ₀/(1 + (r/a)²)²,
where r is the distance to the center, and ρ₀ and a are constants.
(a)  What is the electrical field inside and outside the cylinder?
(b)  Setting V(r=0) = 0, notice the potential at all points r < R.
(c)  What are the equilibrium positions of a particle with accuse q placed inside the cylinder, assuming the charge does not change ρ(r).  What is the forcefulness acting on the particle if information technology is displaced by a distance ε << a from an equilibrium position.  Are the equilibrium positions stable?

Solution:

• Concepts:
  Gauss' law
• Reasoning:
  Choose cylindrical coordinates.  Then E will only have a radial component, E = Eastward(r) (r/r).
• Details of the calculation:
  (a)  For a volume of unit of measurement length:  2πrE = Q_inside/ε₀.
  For r < R:  2πrE = (ρ₀/ε₀)2π∫₀ ^rr'dr'(1 + (r'/a)²)^-ii.
  ε₀rE/ρ₀ =  ∫₀ ^rr'dr'(ane + (r'/a)²)^-2 = aⁱⁱ∫₀ ^r/axdx(1 + xⁱⁱ)^-2 = (a²/two) ∫₀ ^Ydy(1 + y)^-2 = -(a²/2)(i + y)^-1|₀ ^Y
  with Y = (r/a)ⁱⁱ.
  ε₀rE/ρ₀ = (a²/2)(one - ane/(one + (r/a)²)) = (r²/ii) /(1 + (r/a)ⁱⁱ).
  For r < R:  E(r) = ρ₀r/(2ε₀(1 + (r/a)²)),    E(R) = ρ₀R/(2ε₀(1 + (R/a)^two)).
  For r >R:  2πrE = (ρ₀/ε₀)2π∫₀ ^Rr'dr'(1 + (r'/a)ⁱⁱ)^-2.
  For r >R:  E(r) = (ρ₀R²/r)((2ε₀(i + (R/a)²)),  East = abiding/r.

  (b)  V(r < R) = -∫₀ ^rE(r')dr' = -(ρ₀/(2ε₀))∫₀ ^rr'dr'/(ane + (r'/a)²) = -(ρ₀a²/(2ε₀))∫₀ ^r/axdx/(i + x²).
  V(r < R) = -[ρ₀a²/(4ε₀)] ln((r/a)² + ane).

  (c)  All positions on the cylinder axis are equilibrium positions.  With respect to displacements along the cylinder centrality these positions are neutral.  The forcefulness on the particle is zilch.
  With respect to small displacements ε from equilibrium perpendicular to the axis of the cylinder the force is F = qE(r) ≈ qE(0) + q(dE/dr)|₀ε = q(dE/dr)|₀ε = qρ₀ε/(2ε₀) (r/r).
  This is a linear restoring force if q and ρ₀ have opposite signs.
  If q and ρ₀ have reverse signs, then the equilibrium points are stable.
  If q and ρ₀ accept the same sign, and then the equilibrium points are unstable.

---

Gauss' police, planar symmetry

Trouble:

Consider an infinite aeroplane with a uniform accuse density σ located at z = 0.
(a)  Using Gauss' police, find the electric field created by this plane.
(b)  Find the potential Φ(z).
(c)  Locate some other plane with accuse density -σ at z = d.  Observe the potential Φ(z) everywhere.
What is the magnitude of the potential spring across the dipolar layer configurations of the two planes?
(d)  Find the pressure between the planes.

Solution:

• Concepts:
  Gauss' law, the electric potential
• Reasoning:
  We apply Gauss' law to notice the magnitude of the field produced by the plane.
• Details of the calculation:
  (a)  The electric field is perpendicular to the plane by symmetry.
  
  Flux through Gaussian surface = 2E ΔA = σΔA/ε₀
  The field E produced by the canvas is Due east = thou σ/(2ε₀) for z > 0 andDue east = -thousand σ/(2ε₀) for z < 0.

  (b) E = -k ∂Φ/ ∂z.  Allow Φ = 0 at z = 0.  Then Φ = -σ|z|/(2ε₀) for all z.

  (c)  The potential due to the second plate is  Φ = σ|z - d|/(2ε₀) - σd/(2ε₀).
  Using the principle of superposition nosotros have Φ_full = σ|z|/(2ε₀) + σ|z - d|/(2ε₀) + constant.
  If we desire to proceed Φ = 0 at z = 0, then the abiding equals σd/(2ε₀).
  For z < 0:  Φ_total = -σ|z|/(2ε₀) + σ|z|/(2ε₀)  + σd/(2ε₀) - σd/(2ε₀) = 0
  For 0 < z < d:  Φ_total = -σz/(2ε₀) - σz|/(2ε₀)  + σd/(2ε₀) - σd/(2ε₀) =  - σz/ε₀.
  For z > d:  Φ_total = -σz/(2ε₀) + σz|/(2ε₀)  - σd/(2ε₀) - σd/(2ε₀) =  - σd/ε₀.
  The magnitude of the potential jump across the dipolar layer is σd/ε₀.
  (d)  For the two planes: E = -one thousand σ/ε₀ betwixt the planes, Due east = 0 everywhere else.
  The magnitude of the pressure level between the plates is F/A = σ<Eastward> = σ²/(2ε₀)

  Alternative approach:
  U = (ε₀/ii)E² Advert = σ^two/(2ε₀) Advertizing,  |F| = |dU/d(d)| = σ^two/(2ε₀) A,  F/A = σ^two/(2ε₀).

Problem:

A charge Q is placed a distance D from an infinite slab of non-conducting material with charge density ρ and thickness d.  What is the strength on the accuse?

Solution:

• Concepts:
  Gauss' law, F = QE
• Reasoning:
  The problem has enough symmetry to detect the electrical field from Gauss' law lonely.
• Details of the calculation:
  Exterior the slab: Due east _outside = n ρd/(2ε₀), where n is a unit vector perpendicular to the closest surface and pointing away from the surface. F _Q = n Qρd/(2ε₀).

Source: http://electron6.phys.utk.edu/PhysicsProblems/E&M/1-Electrostatics/Gauss.html#:~:text=The%20charge%20distribution%20is%20spherically,%3D%20Qinside%2F%CE%B50.

Posted by: spurgeonfenly1945.blogspot.com

Share this post