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E from Distributions & Gauss' Law Solutions

1. Using the integral form of Coulomb's Law for the following.

(a) Find the electric field at a point a distance a from one end of a long thin wire of length L and total charge Q. Examine the limit a >> L and show that your result is identical to that of a point charge.
(b) If the charge distribution is λ(x) = 2Qx/L², find the electric field. Examine the limit a >> L and show that your result is identical to that of a point charge.



(a) First, the wire has a uniform linear charge density λ = Q/L. Next, we consider a small portion of the wire, dx, located at a distance x from the left end of the wire having charge dq. These infinitesimals are related by

dq = λdx = (Q/L)dx .

Assuming that the charge on the wire is positive, the electric field at x = L + a due to dq, dE, is directed to the right. Using Coulomb's Law, dE is given by

dE = kdq r/r³ .

Here r = i(L+a-x) and r = L+a-x.  Thus

dE = ikdq/r² = i(kQ/L)dx/(L+a-x)² .

Integrating over the entire length of the wire, we get the total electric field to be
 

E	= ∫0L dE
	= i(kQ/L) ∫0L dx/(L+a-x)2
	= i(kQ/L){1/(L+a-x) |0L}
	= i(kQ/L){1/a - 1/(L+a)}
	= i(kQ/a2)(a / L+a)

In the limit a >> L, a/(L+a) ® 1, and

E = i(kQ/a²) .

At a great distance from the wire, the wire produces a field like that of a point charge.

(b) Here we are given the linear charge density λ(x). Next, we consider a small portion of the wire, dx, located at a distance x from the left end of the wire having charge dq. These infinitesimals are related by

dq = λ(x)dx = (2Q/L²) xdx .

Assuming that the charge on the wire is positive, the electric field at x = L + a due to dq, dE, is directed to the right. Using Coulomb's Law, dE is given by

dE = ikdq/r² = i(2Q/L²) dx {x/(L+a-x)²} .

Integrating over the entire length of the wire, we get the total electric field to be
 

E	= ∫0L dE
	= i(2Q/L2) ∫0L dx{x/(L+a-x)2}
	= i(2Q/L2){ (L+a)/(L+a-x) + ln(L + a - x) |0L}
	= i(2Q/L2){L/a - ln(l + L/a)}

In the limit a >> L, {L/a - ln(1 + L/a)} ® ½(L/a)², and

E = i(kQ/a²) .

At a great distance from the wire, the wire produces a field like that of a point charge.



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2. Using the integral form of Coulomb's Law, find the electric field at a point a distance a from the centre of a long thin wire of length L and total charge Q. The identity ∫du[u²+v²]^3/2 = u/{v²[u²+v²]^½} + C may be of use. Show that your result reduces to that of a point charge in the limit a >> L. Also show that your answer reduces to E = 2kλ/a in the limit L >> a, the well-known and very useful result for a long thin wire. Note λ = Q/L.






First, the wire has a uniform linear charge density λ = Q/L. Next, we consider a small portion of the wire, dx, located at a distance x from the left end of the wire having charge dq. These infinitesimals are related by

dq = λdx = (Q/L)dx .

Assuming that the charge on the wire is positive, the electric field at r points away from dq. Using Coulomb's Law, dE is given by

dE = kdq r/r³ .

Here r = &endash;ix + ja and r = (x² + a²)^½.  Thus

 dE = [&endash;ix + ja] (kQ/L)dx/[x² + a²]^3/2 .

Integrating over the entire length of the wire, we get the total electric field to be
 

E	= ∫-½L½L dE
	= (kQ/L) {-i∫-½L½L dx x/[a2 + x2]3/2 + j∫-½L½L dx a/[a2 + x2]3/2}
	= (kQ/L){0 + j x / a[a2+x2]½ |-½L½L }
	= j (kQ/L){½L / a[a2+¼L2]½ - (-½L) / a[a2+¼L2]½}
	= j (kQ/L){2L / a[4a2+L2]½}
	= j (2kQ / La){1 / [1+4(a/L)2]½}

In the limit L >> a, 1 / [1+4(a/L)²]^½® 1, and

E = j(2kQ/La) = j(2kλ/a) .

Note that there is no i component to the electric field as can be seen from the symmetry of the problem.



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3. Using the integral form of Coulomb's Law, find the electric field at a point midway between two long thin wires of length L and total charge Q and off to one side a distance h. The distance between the wires is a. The identity ∫du[u²+v²]^-3/2 = u/{v²[u²+v²]^½} + C may be of use. Examine the limit h >> L and h >> a.






First, each wire has a uniform linear charge density λ = Q/L. Next, we consider a small portion of one wire, dx, located at a distance x from the midpoint of the two wires having charge dq. These infinitesimals are related by

dq = λdx = (Q/L)dx .

Assuming that the charge on the wire is positive, the electric field at r points away from dq. Using Coulomb's Law, dE is given by

dE = kdq r/r³ .

Here r = &endash;ix + jh and r = (x² + h²)^½.  Thus

dE = [&endash;ix + jh](kQ/L)dx/[h² + x²]^3/2 .

The symmetry of the problem can help reduce the amount of work we have to do. First the i components from each wire cancel, so we need only consider the j contributions. Second the j contributions must be the same. As a result we need only integrate over the length of would wire and double the result. We find that the total electric field is
 

E	= 2∫½aL+½a dEy
	= j (2kQ/L) { ∫½aL+½a h/[h2 + x2]3/2 }
	= j (2kQ/L){ x / h[h2+x2]½ |½aL+½a }
	= j (2kQ/L){(L+½a) / h[h2+(L+½a)2]½ - (½a) / h[h2+¼a2]½}

In the limit h >> L and h >> a, (L+½a) / h[h²+(L+½a)²]^½&endash; (½a) / h[h²+¼a²]^½® L/h², and

E = i(2kQ/h²) .

This is same result as from a single point of charge 2Q.



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4. A wire of length 2L is bent in the centre to a right angle. It carries charge density λ. Find an integral expression for the electric field at point P in the upper right corner of the diagram below.
   

   By symmetry, if the electric field due to the bottom half of the wire is E_B = iE_x + jE_y, then the electric field due to the vertical half of the wire is E_V = iE_y + jE_x , i.e. the components flip, and the net electric field would be E_Net = i(E_x + E_y) + j(E_x + E_y). So we need only find the electric field due to the bottom wire to write an expression for the next electric field.

   

   We choose a coordinate system with the bottom wire on the x-axis with the origin at the left side. A tiny piece of the wire dq is chosen at an arbitrary point. The variable of integration is x, and we find dq = λdx, where λ is the linear charge density. The vector from the charge to the field point P is r = i(L – x) + jL, and the magnitude of the vector is r = [(L – x)² + L²]^½. The limits of integration are from x = 0 to x = L.

   The formula for the electric field due to the bottom wire is thus

   E_B = ∫ kdq r/r³ = ∫₀^L kλ dx [i(L – x) + jL]/ [(L – x)² + L²]^3/2

   We immediately can write the electric field due to the vertical wire

   E_V = ∫ kdq r/r³ = ∫₀^L kλ dx [iL + j(L – x)]/ [(L – x)² + L²]^3/2

   The net electric field is thus

   E_Net = E_B + E_V = ∫₀^L kλ dx [–ix – jx]/ [(L – x)² + L²]^3/2



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5. A wire has been bent into a semicircle of radius R. It has a linear charge density λ. Determine the electric field at point P, at the centre of the circle. The identity S = Rθ and dS = Rdθ may help.
   

   We consider a small portion of the wire, dS, located at an angle θ, having charge dq. These infinitesimals are related by

   dq = λdS = λRdθ .

   Assuming that the charge on the wire is positive, the electric field at P points away from dq. Using Coulomb's Law, dE is given by

   dE = kdq r/r

   Here r = –iRcos(θ) –  jRsin(θ) and r = R.  Thus

   dE = [–iRcos(θ) –  jRsin(θ]kλRdθ/R³ = [–icos(θ) –  jsin(θ)] kλdθ/R .

   Integrating over the entire angular length of the wire, we get the total electric field to be

   E	= ∫0π dE
   	= -(kλ/R) { i ∫0πcos(θ)dθ + j ∫0πsin(θ)dθ }
   	= -(kλ/R){ isin(θ)|0π –  jcos(θ)|0π }
   	= -(kλ/R){0 – 2j}
   	= -j(2kλ/R)

   Of course, we could have used the symmetry of the problem to predict that the i component had to be zero.



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6. A thin semicircular wire of radius b has charge Q uniformly distributed along its length. Find an integral expression for the electric field at a point P, a distance a horizontally from its centre.

   

   We choose a coordinate system with the origin of the x-axis at the centre of curvature. A tiny piece of the wire dq is chosen at an arbitrary point. The variable of integration is θ, and we find dq = λ ds = λb dθ where λ is the linear charge density. The vector from the charge to the field point P is r = i(a – b cosθ) – jbsinθ and the magnitude of the vector is r = [(a – bcosθ)² + (bsinθ)²]^½ =  [a² – 2abcosθ + b²]^½. The limits of integration are from θ = –½π to x = ½π.

   The formula for the electric field due to the wire is thus

   E_B = ∫ kdq r/r³ = ∫_–½π^½π kλ dθ [i(a – bcosθ) – j bsinθ] / [a² – 2abcosθ + b²]^3/2



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7. A thin semicircular wire of radius b has charge density +λ on the upper half and opposite charge –λ on the lower half. Find the electric field at the centre of curvature.

   The electric field from the positive half or the arc will point at 45° below the –x-axis which means it will have equal components in the –i and –j directions (see above left). The electric field from the negative half or the arc will point at 45° below the +x-axis which means it will have equal components in the +i and –j directions. For the entire arc, we will be left with just a term that points in the –j direction.  We will work with the positive part of arc and use the symmetry identified to get our final answer (see above right)

   We choose a coordinate system with the origin of the x-axis at the centre of curvature. A tiny piece of the wire dq is chosen at an arbitrary point. The variable of integration is θ , and we find dq = λds = λ b dθ where λ is the linear charge density. The vector from the charge to the field point is r = –ibcosθ – j bsinθ and the magnitude of the vector is r = b. The limits of integration are from θ = 0 to x = ½π

   The formula for the electric field due to the positive portion of the wire is thus

   E₊ = ∫ kdq r/r³ = ∫₀^½π kλ b dθ [–i bcosθ – j bsinθ] / b² = k λ∫₀^½πdθ [–i cos θ – j sinθ] .

   As noted, the symmetry of the problem indicates that the electric field due to the negative portion of the wire is

   E_– = kλ ∫₀^½π dθ [+i cosθ – j sinθ] .

   The net field is thus

   E_Net =  – j 2kλ ∫₀^½π dθ sinθ = – j 2kλ .



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8. Three long thin wires of length L and charges Q, Q, and -Q are arranged to form an equilateral triangle. Use the result of question # 2 to find the electric field at the centre of the triangle.
   

   The magnitude of the electric field from each wire is the same and is directed as shown in the diagram. The net electric field will clearly be downwards, the i contributions will cancel. Using trigonometry, the centre point is distance x = ½Ltan(30°) = L / 2Ö3 from each wire.
   

   Using the result of Question 2,
   

   E_wire = (2kQ / Lx){1 / [1+4(x/L)²]^½} = 6kQ/L² .

   The total electric field is
   

   E_net = -j [E_wire + 2E_wirecos(60°)] = - j 2E_wire = - j 12kQ/L² .

   

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9. The "Gaussian Surface" for an infinitely large charged plate is a pillbox of surface area A and length 2a centred about the origin. The plate has a positive surface charge Σ. (a) How much charge is contained in the pillbox?
   (b) Symmetry demands that the electric field point in the +x direction on the right side of the plate and -x on the other side. Explain why this is so.
   (c) How much electric field passes through the sides of the pillbox?
   (d) The pillbox has a thickness a on either side of the plate. How do the magnitudes of the electric field compare at x = -a and x = +a?
   (e) Use Gauss's Law to determine the electric field at a.
   (f) Sketch the electric field E(x) as a function of x.
   

   (a) The charge contained is Q = A.
   

   (b) A diagram let's us see why the net field is only in the i directions. Consider identical small pieces of charge which are diametrically opposed and the electric field each produces at x = ±a.

   

   If the pieces have the same charge, the electric fields they produce will have the same magnitude. Looking at the above diagram, we see that the j components will cancel.
   

   (c) From part (b) we have seen that the E is in the ±i direction. Therefore no fields pass through the sides of the pillbox as shown in the diagram below.

   

   (d) By symmetry, the electric fields must have the same magnitude on either side of the plate, E(-a) = E(a).

   (e)Gauss' Law states that the total flux of the electric field through all sides of the pillbox is equal to the enclosed charge divided by ε₀, or

   ∫E • ndA = Q_enclosed/ε₀,           (1)

   where n is the outward-looking normal on each side of the surface. From part (a) we know Q_enclosed = A. The pillbox has two ends and one side, so that the integral has three parts
   

   ∫E • ndA = ∫_leftE • n_left dA + ∫_sideE • n_side dA + ∫_rightE • n_right dA .

   Examining the diagram below, we see that E • n_side = 0 since the vectors are at right angles. Similarly, E • n_left = E • n_right = E since the vectors are parallel.
   

   

   Thus equation (1) reduces to
   

   2E ∫_end dA = ΣA/ε₀ .            (2)

   Since E is constant at the ends, it was taken out side the integral. The integral itself then reduces to the surface area of the end of the pillbox,
   

   2EA = ΣA/ε₀ .

   Hence the field due to an infinitely large plate is E = Σ / 2ε₀. This result is good for real plates if one is much closer to the plate than the diameter of the plate.

   

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10. A very long thick plate has a uniform positive volume charge density give by ρ(x) = ρ₀ for -½d £ x £ ½d, where ρ₀ is a positive constant, d is the thickness of the plate and x = 0 is the centre of the plate.
    (a) What is the symmetry of the electric field for this object?
    (b)What is the electric field at x = 0? Explain why.
    (c) How much charge is contained in the a "gaussian pillbox" of surface area A and thickness 2a where a < ½d. Use Gauss' Law to find E(a) .
    (d) How much charge is contained in the a "gaussian pillbox" of surface area A and thickness 2a where a > ½d. Use Gauss' Law to find E(a) .
    (e) Sketch the electric field for all a.

    (a) Symmetry indicates that the electric field only has an i component and no j component. This means that Gauss' Law,

    ∫E • ndA = Q_enclosed/ε₀ ,

    reduces to
    

    2 E(a) A = Q_enclosed/ε₀ .

    (b) The electric field at x = 0 is zero since the charge is distributed symmetrically about the y-axis. The electric field from a piece of charge on one side will exactly cancel with the field from a piece of charge on the other side.
    

    (c) We need to find the charge inside the pillbox shown in the diagram below.
    

    

    The charge inside is given by
    

    Q_enclosed = ρ₀V = ρ₀(2a)A = 2aAρ₀ .

    Thus Gauss' Law yields,

    2 E(a) A = 2aAρ₀/ε₀ .

    Thus the electric field is

    E(a) = ρ₀a/ε₀ .

    (d) We need to find the charge inside the pillbox shown in the diagram below.
    

    

    The charge inside is given by
    

    Q_enclosed = ρ₀V = ρ₀Ad ,

    since the charge is confined to -½d £ x £ ½d.

    Thus Gauss' Law yields,

    2 E(a) A = Adρ₀/ρ₀.

    Thus the electric field is E(a) = ρ₀d / 2ε₀ .

    (e) The electric field as a function of a looks like

    

    

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11. The "Gaussian Surface" for cylinders is also a cylinder. It has length L and radius a. Consider an infinitely long wire of uniform charge per unit length λ.
    (a) How much electric field passes through the ends of the "gaussian cylinder". Explain.
    (b) How much charge is contains in the "gaussian cylinder"?
    (c) Use Gauss's Law to determine the electric field at a distance a.
    

    The amount of flux that passes through the cylinder is given by

    ∫E • ndA = ∫_leftE • n_left dA + ∫_sideE • n_side dA + ∫_rightE • n_right dA .

    Examining the diagram above, we see that E • n_side = E(a) since the vectors are parallel. Similarly, E • n_left = E • n_right = 0 since the vectors are at right angles. So there is no flux through the ends of the "gaussian cylinder".
    

    (b) Only a length L of the wire is inside the "gaussian cylinder", so Q_enclosed = λL .
    

    (c) Gauss' Law states that the total flux of the electric field through all sides of the pillbox is equal to the enclosed charge divided by ε₀, or

    ∫E • ndA = Q_enclosed/ε₀ ,

    where n is the outward-looking normal on each side of the surface. From parts (a) and (b), we know that this reduces to
    

    ∫_side E(a)dA = λL/ε₀ .

    Since E(a) is a constant for a given radius a, it may be taken outside the integral,
    

    E(a) ∫_side dA = λL/ε₀ ,

    and we are left with an integral for the surface area of the side of the "gaussian cylinder" which is A = 2πaL. Thus we have
    

    E(a) 2πaL = λL/ε₀ .

    Solving for the electric field yields

    E(a) = λ / 2pe₀a .

    

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12. A very long thin cylindrical shell of radius R carries a negative surface charge density Σ.
    (a) If the radius of the "gaussian cylinder" is smaller than R, i.e. a < R, how much charge is contained by the gaussian surface?
    (b) What, therefore, is the electric field everywhere inside the cylindrical shell?
    (c) How much charge is contained in the "gaussian cylinder" when a > R?
    (d) Use Gauss's Law to determine the electric field at a.
    (e) Sketch the electric field as a function of a.
    

    For cases of cylindrical symmetry, Gauss' Law
    

    ∫E • ndA = Q_enclosed/ε₀ ,

    reduces to the much simpler
    

    E(a) 2πaL = Q_enclosed/ε₀ ,

    where a is the radius of the "gaussian cylinder" and L is its length.
    

    (a) We draw an end view of the shell and "gaussian cylinder".
    

    

    We see that there is no charge inside the "gaussian cylinder", so Q_enclosed = 0.
    

    (b) Thus we find that the electric field for a < R is 0. There is no electric inside the shell.
    

    (c) Considering the case of a > R, we have the following diagram.
    

    

    The "gaussian cylinder" encircles a length L of the shell, so Q_enclosed = Σ2πRL.
    

    (d) Thus Gauss' Law reduces to

    E(a) 2πaL = Σ2πRL / ε₀ ,

    and hence the electric field is
    

    E(a) = ΣR / aε₀ ,

    for a > R.
    

    (e) The electric field as a function of a looks like

    

    Note the jump in the electric field at the shell.
    

    

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13. A solid sphere of radius R carries a total positive charge Q uniformly distributed throughout the sphere. The "Gaussian Surface" for a sphere is a sphere of radius a concentric with the sphere.
    (a) What is the electric field at the centre of the sphere? Explain?
    (b) If a < R, how much charge is contained inside the "gaussian sphere"?
    (c) Use Gauss's Law to determine the electric field at the surface of the "gaussian sphere".
    (d) If a > R, how much charge is contained inside the "gaussian sphere"?
    (e) Use Gauss's Law to determine the electric field at the surface of the "gaussian sphere".
    (f) Sketch the electric field.
    

    (a) The charge is arranged symmetrically about the centre, so the electric field from a piece of charge on one side is cancelled by the electric field from the diametrically opposed piece of charge on the other side. Thus the electric field is zero at the centre.
    

    (b) We sketch the problem

    

    The "gaussian sphere" contains only a portion of the total charge given by
    

    Q_enclosed = Q V_gauss/V_sphere = Q (4πa³/3 / 4πR³/3) = Q(a/R)³ .

    (c) Gauss' Law is stated
    

    ∫E • ndA = Q_enclosed/ε₀ ,

    where n is the outward looking normal. For cases of spherical symmetry, as we have here, E and n are parallel and E • n = E(a). The electric field at the surface, E(a), is a constant because of the spherical charge distribution and may be taken outside the integral. We determined Q_enclosed in part (b), thus we have

    E(a) ∫ dA = (Q/ε₀)(a/R)³ .

    The integral is simply the surface area of the "gaussian sphere", hence we have
    

    E(a) 4πa² = (Q/ε₀)(a/R)³ .

    Thus the electric field is
    

    E(a) = Qa / 4pe₀R³ ,

    for 0 £ a £ R.
    

    (d) We sketch the "gaussian sphere" and see that it contains the entire sphere therefore Q_enclosed = Q.
    

    

    (e) Gauss' Law reduces to
    

    E(a) 4πa² = Q/ε₀ ,

    and the electric field is
    

    E(a) = Q / 4pe₀a² ,

    for a ³ R.
    

    (f) The electric field looks like

    

    

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14. A spherical shell of inner radius R and outer radius 2R, has a uniform charge distribution and total charge Q.
    (a) Determine the charge inside the "gaussian sphere" for the three regions
      (i)   0 < a < R,
      (ii)  R < a < 2R,
      (iii) 2R < a.
    (b) Use Gauss's Law to determine the electric field at the surface of the "gaussian surface" when
      (i)   0 < a < R,
      (ii)  R < a < 2R,
      (iii) 2R < a.
    (c) Sketch the electric field for all a.
    

    (a) First we sketch the "gaussian sphere" in each case.
    

    
    

    (i) Here we see that there is no charge inside the "gaussian sphere", so Q_enclosed = 0.
    

    (ii) Here there is only a portion of the total charge inside the "gaussian sphere" given by the ratio of volumes

    

    (iii) Here the entire shell is enclosed, so Q_enclosed = Q.
    

    (b) In cases of spherical symmetry, Gauss' Law
    

    ∫E • ndA = Q_enclosed/ε₀ ,

    reduces to
    

    E(a) 4πa² = Q_enclosed / ε₀ .

    For our three cases, we have

    (i) For a £ R,

    E(a) 4πa² = 0 .

    Thus E(a) = 0.
    

    (ii) For R £ a £ 2R,
    

    E(a) 4πa² = Q(a³-R³)/(7ε₀R³) .

    Thus E(a) = (Q/4pe₀R³)(a³-R³)/(7a²).

    (iii) For a £ 2R,

    E(a) 4πa² = Q/ε₀ .

    Thus E(a) = Q/4pe₀a².
    

    (c) The electric field looks like

    



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15. Use Gauss's Law to determine the electric field as a function of distance a from the centre of a thin spherical shell of radius R and negative surface charge -Σ. Sketch the electric field as a function of a. By looking at the electric field, can we distinguish between a sphere with all its charge on the surface and a sphere with the charge smeared throughout the sphere?
    

    For cases of spherical symmetry, Gauss' Law
    

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     <hr><a name="Q15"></a>
      <li>Use
    ∫E • ndA = Q_enclosed/ε₀ ,

    reduces to
    

    E(a) 4πa² = Q_enclosed/ε₀ .

    Here we have two cases as shown in the diagram below: (i) 0 £ a £ R and (ii) a ³ R.
    

    

    (i) For 0 £ a £ R, Q_enclosed = 0. Hence the electric field is zero.
    

    (ii) For a ³ R, the total shell is enclosed so Q_enclosed = Q. We are not given Q but rather the charge density so

    Q_enclosed = -Σ 4πR² .

    Thus Gauss' Law is
    

    E(a) 4πa² = -Σ 4πR² / ε₀ .

    Hence the electric field is
    

    E(a) = -(Σ/ε₀)(R/a)² .

    Thus the electric field looks like
    

    

    As long as the charge distribution is symmetrical, we can find the electric field from Gauss' Law. Gauss' Law says that E(a) depends only on the total charge enclosed and thus is independent of how that charge is distributed throughout the material. Hence we cannot distinguish between a sphere with all its charge on the surface and a sphere with the charge smeared throughout the sphere.



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16. An object consists of two thin spherical shells or radius R and 2R respectively (see diagram below). On the inner shell there is a charge Q and on the outer shell there is a charge –Q. Find and sketch the electric field everywhere.

    

    We have spherical symmetry so we use spherical Gaussian surfaces with the same centre. There are three different regions.

    

    In the first region, 0 ≤ r < R, nothing is enclosed. So Q_enclosed = 0 and thus E = 0.

    In the second region, R < r < 2R, only the innermost shell and its charge are enclosed. So Q_enclosed = Q. For spherical symmetry Gauss's Law is E4πr² = Q_enclosed/ε₀. So E = Q/4ε₀r².

    In the third region, r > 2R, both shells are enclosed. So Q_enclosed = Q + –Q = 0 and thus E = 0.

    The electric field of E vs r looks like the graph below. The values on the y-axis are found by setting r = R and r = 2R in the equation for E in the region R < r < 2R. Note the jumps because of the surface charges (i.e. shells).

    



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17. An object consists of two thin spherical shells or radius R and 2R respectively (see diagram above). On the inner shell there is charge density σ and on the outer shell there is charge density –σ. What is the charge on each shell? Find and sketch the electric field everywhere. What does the charge on the outside shell have to be for E(r > 2R) = 0?

    Surface charge density is defined σ = Q/A. The surface area of a sphere is 4πr². Thus the charge on the inner shell is Q₁ = σ4πR² and the charge on the outer shell is Q₂ = –σ4π(2R)² = –σ16πR² = –4Q₁.

    /p>

    As in the previous problem, we have spherical symmetry so we use spherical Gaussian surfaces with the same centre. There are the same three regions.

    In the first region, 0 ≤ r < R, nothing is enclosed. So Q_enclosed = 0 and thus E = 0.

    In the second region, R < r < 2R, only the innermost shell and its charge are enclosed. So Q_enclosed = Q₁. For spherical symmetry Gauss's Law is E4πr² = Q_enclosed/e₀. So E = σ4πR²/4πε₀r² or E = σR²/ε₀r²

    In the third region, r > 2R, both shells are enclosed. So Q_enclosed = Q₁ + Q₂ = –3Q₁ and thus E = 3σ4πR²/4πε₀r² or E = –3σR²/ε₀r².

    The electric field of E vs r looks like the graph below. The values on the y-axis are found by setting r = R and r = 2R in the equation for E in the region R < r < 2R and r = 2R for the field in the outermost region. Note the jumps because of the surface charges (i.e. shells).

    



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18. A very long cylindrical object consists of a solid core of radius R and a thin shell at radius 2R. The core has linear charge density λ. The shell has linear charge density –λ. What is the charge of a piece of the core of length L? What is the charge of a piece of the shell length L? Find and sketch the electric field everywhere.

    

    A linear charge density is λ = Q/L so Q₁ = λL for the inner core and Q₂ = –λL for the outer shell. The charge density and hence the field is cylindrical so we use a Gaussian cylinder of length L with a common centre with the object. Looking end on, we have three regions.

    /p>

    In the first region, 0 ≤ r < R, only part of the core is enclosed. We find out by noting that Q_enclosed = Q₁ × V_enclosed/V₁ = λL(πr²L/πR²L) = λLr²/R². For cylindrical symmetry, E2πrL = Q_enclosed/ε₀. So E = λLr²/R² / ε₀2πrL = λr/2πε₀R².

    In the second region, R < r < 2R, the core is enclosed so Q_enclosed = Q₁ = λL. For cylindrical symmetry, E2πrL = Q_enclosed/e₀. So E = λL / ε₀2πrL = λ/2πε₀r.

    In the third region, r > 2R, both the core and the outer shell are enclosed. Thus Q_enclosed = Q₁ + Q₂ = λL + –λL = 0. So E = 0.

    To graph the electric field appropriately, note the value of E at the boundaries of each region. Note the kink where the solid ends and the drop at the shell.

    



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19. An object consists of three very large parallel planes of material. The gap between the planes is distance d.  The planes have surface charge densities σ, –σ, and σ from right to left. The planes are shown edge on below. What is the charge of a square of area A of each of these planes? Find and sketch the electric field everywhere. Suppose the planes have surface charge densities σ, 2σ, and 3σ from right to left. Could you still find the electric field? Why or why not?

    

    The charge on area A of the positive plate is Q₊ = σA while the negative plate has charge Q_– = –σA.

    The symmetry is planar so we use a Gaussian pillbox of end area A centered on the plates. There are only two ways to do this.

    

    In the first region, 0 ≤ x < d, only part of the centre plate is enclosed and Q_enclosed = Q_– = –σA. For planar symmetry, Gauss`s Law reduces to E(2A) = Q_enclosed/e₀. So E = –σ/2ε₀.

    In the second region, x > d, part of all three plates are enclosed and Q_enclosed = Q₊ + Q_– + Q₊ = σA. So E(2A) = Q_enclosed/ε₀ yields E = σ/2ε₀.

    The electric field graph looks like:

    

    If the charge densities were σ, 2σ, and 3σ, we could not use Gauss`s Law directly because the electric field would not be symmetric. E would be different at each end of the pillbox.

    Despite this we could use Gauss`s Law if we recall that the net electric field is the sum of the individual fields and they have the required planar symmetry. So find the field for each plate separately, then add carefully.



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20. A conducting hollow spherical shell carries a net charge of +5Q. A charge of –2Q in placed in the hollow. What will the charge be on each surface of the shell after the charge is put inside?

    We have several concepts that allow us to solve this sort of question.

    • the electric field in a conductor is zero in equilibrium
    • charge, induced or added, can only reside on interior or exterior surfaces
    • charge is conserved
    • flux through a Gaussian surface depends only on the enclosed charge

    

    The original +5Q must be on the exterior of the spherical shell as picture left above. We draw a Gaussian surface just inside the conducting shell that now contains the charge (see above middle). The flux through this surface must be zero since the electric field inside a conductor is zero. The flux is zero only if the enclosed charge is zero. We certainly have –2Q inside so +2Q must have been induced on the interior surface. Where did the +2Q come from? It came from the exterior surface, which leaves +3Q on the exterior surface. See the above picture on the right for the final charge distribution.

    

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21. A charge of –2Q in placed in a conducting box which is then placed inside a conducting hollow cylinder. The box was initially uncharged but the cylinder already carried a charge of +2Q. Find the charge on each surface.
    

    We have several concepts that allow us to solve this sort of question.

    • the electric field in a conductor is zero in equilibrium
    • charge, induced or added, can only reside on interior or exterior surfaces
    • charge is conserved
    • flux through a Gaussian surface depends only on the enclosed charge

    The shapes of the conductors do not matter. When the –2Q charge is placed in the box, a charge +2Q is induced on the interior surface of the box to keep the field in the box zero. Since the box was neutral that means there has to be –2Q on the exterior surface of the box to compensate. When the charge and box are placed in the cylinder, a charge of +2Q is induced on the interior surface of the cylinder to keep the electric field in the conducting shell of the cylinder zero. Since the cylinder originally had a net charge of +2Q, all the exterior charge moved to the interior surface leaving no charge on the exterior surface. The results are summarized in the drawing below.

    

Questions? mike.coombes@kwantlen.ca